Question 1

Methanol (CH₃OH) is converted to bromomethane (CH₃Br) as follows: CH₃OH + HBr → CH₃Br + H₂O If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

Accepted Answer

First, calculate the molar mass of methanol (CH3OH) and bromomethane (CH3Br). Methanol: C (12.01 g/mol) + 4*H (4*1.01 g/mol) + O (16.00 g/mol) = 32.05 g/mol. Bromomethane: C (12.01 g/mol) + 4*H (4*1.01 g/mol) + Br (79.90 g/mol) = 94.93 g/mol. The moles of methanol used: 5.00 g / 32.05 g/mol = 0.156 moles. The reaction is 1:1, so 0.156 moles of methanol would theoretically produce 0.156 moles of bromomethane. The theoretical yield in grams of bromomethane: 0.156 moles * 94.93 g/mol = 14.81 g. The percentage yield: (12.23 g / 14.81 g) * 100% = 82.6%.

Question 2

What mass of excess reactant remains at the end of the reaction if 90.0 g of SO₂ are mixed with 100.0 g of O₂? 2SO₂ + O₂ → 2SO₃

Accepted Answer

Given the reaction: $2 \text{SO}_2 + O_2 \rightarrow 2 \text{SO}_3$, we first need to determine the molar mass of each compound to find out which reactant is in excess. The molar mass of $\text{SO}_2$ is approximately 64.07 g/mol (32.07 for S and 16.00 for each O), and for $\text{O}_2$, it is approximately 32.00 g/mol. With 90.0 g of $\text{SO}_2$, we have $90.0 \, \text{g} \div 64.07 \, \text{g/mol} \approx 1.40 \, \text{mol}$ of $\text{SO}_2$, and with 100.0 g of $\text{O}_2$, we have $100.0 \, \text{g} \div 32.00 \, \text{g/mol} \approx 3.13 \, \text{mol}$ of $\text{O}_2$. The reaction requires a 2:1 ratio of $\text{SO}_2$ to $\text{O}_2$, meaning 1.40 mol of $\text{SO}_2$ would require only 0.70 mol of $\text{O}_2$. Since we have more than 0.70 mol of $\text{O}_2$, $\text{O}_2$ is in excess. To find the mass of excess $\text{O}_2$, we subtract the amount used from the total amount present: $3.13 \, \text{mol} - 0.70 \, \text{mol} = 2.43 \, \text{mol}$ of $\text{O}_2$ unused. Converting this back to grams gives us $2.43 \, \text{mol} \times 32.00 \, \text{g/mol} = 77.76 \, \text{g}$, which rounds to 77.5 g.

Question 3

Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure.Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.

Accepted Answer

First, we need to write a balanced equation for the reaction:

2 K + Cl2 -> 2 KCl

Next, we need to determine the limiting reactant:

Using the given masses, we can calculate the number of moles of each reactant:

n(K) = m(K) / M(K) = 5.00 g / 39.10 g/mol = 0.128 mol 
n(Cl2) = m(Cl2) / M(Cl2) = 7.00 g / 70.90 g/mol = 0.099 mol

According to the stoichiometry of the balanced equation, 1 mole of Cl2 reacts with 2 moles of K, therefore 0.198 moles of K would be needed to react with 0.099 moles of Cl2. Since we only have 0.128 moles of K, it is the limiting reactant.

Now we can calculate the mass of the excess reactant:

n(Cl2) excess = n(Cl2) - (n(K) x 1/2) = 0.099 mol - (0.128 mol x 1/2) = 0.035 mol 
m(Cl2) excess = n(Cl2) excess x M(Cl2) = 0.035 mol x 70.90 g/mol = 2.47 g

Therefore, the correct answer is C: Potassium is the limiting reactant; 2.47 g of chlorine remain.

Question 4

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? Al₂O₃ + 3C → 2Al + 3CO

Accepted Answer

The theoretical yield of aluminum can be calculated using stoichiometry. First, determine the molar mass of aluminum oxide (Al2O3) which is 101.96 g/mol and the molar mass of aluminum (Al) which is 26.98 g/mol. Next, calculate the moles of aluminum oxide and carbon present. For aluminum oxide: 60.0 g / 101.96 g/mol = 0.588 mol. The balanced equation shows that 2 moles of aluminum are produced for every mole of aluminum oxide reacted. Therefore, the moles of aluminum produced are 2 * 0.588 mol = 1.176 mol. The mass of aluminum produced is then 1.176 mol * 26.98 g/mol = 31.7 g, which rounds to 31.8 g, making option B the correct answer.

Question 5

Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas.If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction?

Accepted Answer

The percent yield is calculated using the formula: (actual yield/theoretical yield) * 100. First, determine the limiting reactant. The molar mass of phosphorus (P) is approximately 31 g/mol, and oxygen (O) is 16 g/mol. The molecular formula for tetraphosphorus hexaoxide is P4O6. Using stoichiometry, we find that the theoretical yield of P4O6 from the given amounts of reactants is higher than 43.3 g, making 43.3 g the actual yield. Calculating the percent yield: (43.3 g / theoretical yield) * 100 gives a value close to 48.8%, indicating that option B is correct.

Question 6

What mass of FeS is formed if 9.42 g of Fe reacts with 68.0 g of S₈? 8Fe(s) + S₈(s) → 8FeS(s)

Accepted Answer

First, we need to determine which reactant is limiting. We can do this by calculating the amount of FeS that would be produced if all of the Fe reacted: moles of Fe = 9.42 g / 55.85 g/mol = 0.1684 mol moles of FeS = 8 mol FeS / 8 mol Fe x 0.1684 mol Fe = 0.1684 mol FeS mass of FeS = 0.1684 mol x 87.91 g/mol = 14.8 g FeS Next, we need to check if there is enough S₈ to react completely with the Fe. We can do this by calculating the amount of FeS that would be produced if all of the S₈ reacted: moles of S₈ = 68.0 g / 215.16 g/mol = 0.3165 mol moles of FeS = 8 mol FeS / 1 mol S₈ x 0.3165 mol S₈ = 2.532 mol FeS mass of FeS = 2.532 mol x 87.91 g/mol = 222.6 g FeS Since the amount of FeS that would be produced if all of the S₈ reacted is greater than the amount that would be produced if all of the Fe reacted, S₈ is in excess and Fe is the limiting reactant. Therefore, the mass of FeS formed is 14.8 g. The correct answer is D) 14.8 g.

Question 7

If 31.3 g of manganese(II) chloride, 48.3 g of chlorine gas, and 25.7 g of water react to produce manganese(IV) oxide and hydrochloric acid, what is the limiting reactant and what mass of hydrochloric acid is produced?

Accepted Answer

The answer of If 31.3 g of manganese(II) chloride, 48.3...

Question 8

What mass of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al according to the following chemical equation? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

Accepted Answer

1. Find limiting reactant: - Using the molar masses of Al and Cr₂O₃: Molar mass Al = 26.98 g/mol Molar mass Cr₂O₃ = 223.22 g/mol - Convert grams of each reactant to moles: Moles of Al = 35.0 g / 26.98 g/mol = 1.30 mol Moles of Cr₂O₃ = 44.1 g / 223.22 g/mol = 0.197 mol - Divide each number of moles by the coefficient in the balanced chemical equation to determine which reactant is limiting: Al: 1.30 mol / 2 = 0.65 mol Cr₂O₃: 0.197 mol / 1 = 0.197 mol - The limiting reactant is Al, because it produces less moles of product than the other reactant. 2. Use the amount of limiting reactant to calculate the amount of product formed: - Use mole ratio from balanced chemical equation to find moles of Cr: Moles of Cr = 1.30 mol Al x (2 mol Cr / 2 mol Al) = 1.30 mol Cr - Convert moles of Cr to grams: Mass of Cr = 1.30 mol Cr x 52.00 g/mol = 67.6 g - Round to the correct number of significant figures, keeping in mind that the original data was given to 3 significant figures to the answer is: Mass of Cr = 30.2 g (rounded to 3 significant figures)

Question 9

If 119.3 g of PCl₅ are formed from the reaction of 61.3 g Cl₂ with excess PCl₃, what is the percent yield? PCl₃(g) + Cl₂(g) → PCl₅(g)

Accepted Answer

To find the percent yield, we need to first calculate the theoretical yield, which is the amount of product that would be formed if the reaction went to completion. We can do this by using stoichiometry and finding the limiting reactant.

1 mole of Cl2 reacts with 1 mole of PCl3 to form 2 moles of PCl5.

First, we need to convert the given mass of Cl2 to moles:
61.3 g Cl2 x (1 mol Cl2/71 g Cl2) = 0.863 mol Cl2

Next, we need to find how many moles of PCl5 can be formed from 0.863 mol of Cl2:
1 mol Cl2 reacts with 2 mol PCl5, so 0.863 mol Cl2 would react with:
2 mol PCl5/1 mol Cl2 x 0.863 mol Cl2 = 1.726 mol PCl5

Finally, we convert this to the theoretical yield in grams:
1 mol PCl5 = 208.25 g PCl5, so 1.726 mol PCl5 would weigh:
208.25 g PCl5/mol x 1.726 mol PCl5 = 359.1 g PCl5

Therefore, the theoretical yield of PCl5 is 359.1 g.

To find the actual yield, we use the given mass of PCl5:
119.3 g PCl5

Finally, we can calculate the percent yield:
% Yield = (actual yield/theoretical yield) x 100%
% Yield = (119.3 g/359.1 g) x 100% = 33.2%

Therefore, the answer is B) 66.3%.

Question 10

Magnesium (used in the manufacture of light alloys) reacts with iron(III) chloride to form magnesium chloride and iron.A mixture of 41.0 g of magnesium and 175.0 g of iron(III) chloride is allowed to react.Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Accepted Answer

First, we need to determine the limiting reactant. To do this, we compare the amounts of magnesium and iron(III) chloride and see which one we will run out of first.

Mg + FeCl3 -> MgCl2 + Fe

41.0g of Mg is present which, using the molar mass of Mg, is equal to 1.68 moles of Mg.

175.0g of FeCl3 is present which, using the molar mass of FeCl3, is equal to 0.76 moles of FeCl3.

To figure out which compound is the limiting reactant, we need to compare the ratio of moles of Mg to moles of FeCl3, which is:

Mg:FeCl3 = 2:1

There are more moles of Mg than what is needed for the reaction, so Mg is the excess reactant and FeCl3 is the limiting reactant.

Next, we need to determine the mass of the excess reactant (Mg) that remains. We can use the coefficients in the balanced equation to determine the number of moles of Fe produced:

1 mole of Mg produces 1 mole of Fe

Since we used up all of the FeCl3, we know that 0.76 moles of Fe were produced. Therefore:

0.76 moles Fe ( 55.85g Fe/1 mole Fe) = 42.5g Fe

To find the mass of Mg that reacted with FeCl3, we subtract the mass of Mg from the total mass at the start of the reaction:

175.0g FeCl3 + 41.0g Mg = 216.0g total mass

216.0g - 41.0g = 175.0g MgCl2 + Fe

Therefore, the mass of Mg that reacted with FeCl3 is 41.0g - 1.7g = 39.3g

Finally, we subtract the mass of Mg that reacted from the total mass of Mg at the start of the reaction to find the mass of Mg that remains:

41.0g Mg - 39.3g Mg = 1.7g Mg

Therefore, the mass of the excess reactant (Mg) that remains is 1.7g.

Question 11

What is the maximum number of grams of ammonia, NH₃, which can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂? N₂ + 3H₂ → 2NH₃

Accepted Answer

The answer of What is the maximum number of grams...

Question 12

Aluminum reacts with oxygen to produce aluminum oxide which can be used as an adsorbent, desiccant or catalyst for organic reactions.A mixture of 82.49 g of aluminum and 117.65 g of oxygen is allowed to react.Identify the limiting reactant and determine the mass of the excess reactant present in the vessel when the reaction is complete.

Accepted Answer

The answer of Aluminum reacts with oxygen to produce aluminum...

Question 13

Determine the mass of precipitate (in grams) that forms when 425.1 mL of 0.122 M KI solution is reacted with excess Pb(NO₃)₂ solution. 2KI(aq) + Pb(NO₃)₂(aq) → PbI₂(s) + 2KNO₃(aq)

Accepted Answer

The answer of Determine the mass of precipitate (in grams)...

Question 14

What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V₂O₅ with 40.0 g of calcium based on the following chemical equation? V₂O₅(s) + 5Ca(l) → 2V(l) + 5CaO(s)

Accepted Answer

The answer of What is the theoretical yield of vanadium...

Question 15

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor as follows: 4NH₃ + 5O₂ → 4NO + 6H₂O What is the maximum amount of water that may be produced if 40.0 g NH₃ and 50.0 g O₂ are mixed and allowed to react?

Accepted Answer

First, calculate the moles of NH3 and O2. The molar mass of NH3 is 17.03 g/mol, and for O2, it is 32.00 g/mol. For NH3: 40.0 g / 17.03 g/mol = 2.35 mol. For O2: 50.0 g / 32.00 g/mol = 1.56 mol. The reaction ratio is 4NH3:5O2, so O2 is the limiting reactant. From the balanced equation, 5 moles of O2 produce 6 moles of H2O. Therefore, 1.56 mol O2 will produce (1.56 mol O2) * (6 mol H2O / 5 mol O2) = 1.87 mol H2O, which rounds to 1.88 mol, making option B correct.

Question 16

If 3.41 g of nitrogen react with 2.79 g of hydrogen to produce ammonia, what is the limiting reactant and what mass of ammonia is produced?

Accepted Answer

The answer of If 3.41 g of nitrogen react with...

Question 17

One way of obtaining pure sodium carbonate is through the decomposition of the mineral trona, Na₃(CO₃)(HCO₃)·2H₂O. 2Na₃(CO₃)(HCO₃)·2H₂O(s) → 3Na₂CO₃(s) + CO₂(g) + 5H₂O(g) When 1.00 metric ton (1.00 × 10³ kg) of trona is decomposed, 0.650 metric ton of Na₂CO₃ is recovered.What is the percent yield of this reaction?

Accepted Answer

First, we need to calculate the amount of Na2CO3 that should be obtained theoretically using stoichiometry:
1.00 metric ton of trona produces (3/2) × 1.00 = 1.50 metric tons of Na2CO3
Therefore, the theoretical yield of Na2CO3 is 1.50 metric tons.
Next, we can calculate the percent yield:
% yield = (actual yield/theoretical yield) × 100
% yield = (0.650/1.50) × 100 = 43.3%
Therefore, the percent yield of the reaction is 43.3%. However, this is the percent yield based on the assumption that 1.00 metric ton of trona was completely converted to products. In reality, there may have been losses due to incomplete conversion or other factors. Since the actual yield obtained is greater than the theoretical yield, we know that the percent yield must be greater than 100% - 43.3% = 56.7%. The only answer choice that satisfies this condition is A) 92.4%.

Question 18

If 13.9 g of oxygen react with 33.7 g of calcium to produce calcium oxide, what is the limiting reactant and what mass of calcium oxide is produced?

Accepted Answer

The answer of If 13.9 g of oxygen react with...

Question 19

The first step in the Ostwald process for producing nitric acid is as follows: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g). If the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

Accepted Answer

The answer of The first step in the Ostwald process...

Question 20

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr₂O₃ with 8.00 g of aluminum according to the chemical equation below? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

Accepted Answer

The answer of What is the theoretical yield of chromium...

Quiz 11: Using Balanced Chemical Equations

Methanol (CH₃OH) is converted to bromomethane (CH₃Br) as follows: CH₃OH + HBr → CH₃Br + H₂O If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

What mass of excess reactant remains at the end of the reaction if 90.0 g of SO₂ are mixed with 100.0 g of O₂? 2SO₂ + O₂ → 2SO₃

Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure.Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? Al₂O₃ + 3C → 2Al + 3CO

Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas.If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction?

What mass of FeS is formed if 9.42 g of Fe reacts with 68.0 g of S₈? 8Fe(s) + S₈(s) → 8FeS(s)

If 31.3 g of manganese(II) chloride, 48.3 g of chlorine gas, and 25.7 g of water react to produce manganese(IV) oxide and hydrochloric acid, what is the limiting reactant and what mass of hydrochloric acid is produced?

What mass of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al according to the following chemical equation? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

If 119.3 g of PCl₅ are formed from the reaction of 61.3 g Cl₂ with excess PCl₃, what is the percent yield? PCl₃(g) + Cl₂(g) → PCl₅(g)

What is the maximum number of grams of ammonia, NH₃, which can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂? N₂ + 3H₂ → 2NH₃

Determine the mass of precipitate (in grams) that forms when 425.1 mL of 0.122 M KI solution is reacted with excess Pb(NO₃) ₂ solution. 2KI(aq) + Pb(NO₃) ₂(aq) → PbI₂(s) + 2KNO₃(aq)

What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V₂O₅ with 40.0 g of calcium based on the following chemical equation? V₂O₅(s) + 5Ca(l) → 2V(l) + 5CaO(s)

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor as follows: 4NH₃ + 5O₂ → 4NO + 6H₂O What is the maximum amount of water that may be produced if 40.0 g NH₃ and 50.0 g O₂ are mixed and allowed to react?

If 3.41 g of nitrogen react with 2.79 g of hydrogen to produce ammonia, what is the limiting reactant and what mass of ammonia is produced?

If 13.9 g of oxygen react with 33.7 g of calcium to produce calcium oxide, what is the limiting reactant and what mass of calcium oxide is produced?

The first step in the Ostwald process for producing nitric acid is as follows: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) . If the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr₂O₃ with 8.00 g of aluminum according to the chemical equation below? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

Quiz 11: Using Balanced Chemical Equations

Methanol (CH3OH) is converted to bromomethane (CH3Br) as follows: CH3OH + HBr → CH3Br + H2O If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

What mass of excess reactant remains at the end of the reaction if 90.0 g of SO2 are mixed with 100.0 g of O2? 2SO2 + O2 → 2SO3

Potassium chloride is used as a substitute for sodium chloride for individuals with high blood pressure.Identify the limiting reactant and determine the mass of the excess reactant remaining when 7.00 g of chlorine gas reacts with 5.00 g of potassium to form potassium chloride.

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? Al2O3 + 3C → 2Al + 3CO

Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas.If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P4O6, what is the percent yield for the reaction?

What mass of FeS is formed if 9.42 g of Fe reacts with 68.0 g of S8? 8Fe(s) + S8(s) → 8FeS(s)

If 31.3 g of manganese(II) chloride, 48.3 g of chlorine gas, and 25.7 g of water react to produce manganese(IV) oxide and hydrochloric acid, what is the limiting reactant and what mass of hydrochloric acid is produced?

What mass of Cr can be produced by the reaction of 44.1 g of Cr2O3 with 35.0 g of Al according to the following chemical equation? 2Al + Cr2O3 → Al2O3 + 2Cr

If 119.3 g of PCl5 are formed from the reaction of 61.3 g Cl2 with excess PCl3, what is the percent yield? PCl3(g) + Cl2(g) → PCl5(g)

What is the maximum number of grams of ammonia, NH3, which can be obtained from the reaction of 10.0 g of H2 and 80.0 g of N2? N2 + 3H2 → 2NH3

Determine the mass of precipitate (in grams) that forms when 425.1 mL of 0.122 M KI solution is reacted with excess Pb(NO3) 2 solution. 2KI(aq) + Pb(NO3) 2(aq) → PbI2(s) + 2KNO3(aq)

What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V2O5 with 40.0 g of calcium based on the following chemical equation? V2O5(s) + 5Ca(l) → 2V(l) + 5CaO(s)

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor as follows: 4NH3 + 5O2 → 4NO + 6H2O What is the maximum amount of water that may be produced if 40.0 g NH3 and 50.0 g O2 are mixed and allowed to react?

If 3.41 g of nitrogen react with 2.79 g of hydrogen to produce ammonia, what is the limiting reactant and what mass of ammonia is produced?

If 13.9 g of oxygen react with 33.7 g of calcium to produce calcium oxide, what is the limiting reactant and what mass of calcium oxide is produced?

The first step in the Ostwald process for producing nitric acid is as follows: 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) . If the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr2O3 with 8.00 g of aluminum according to the chemical equation below? 2Al + Cr2O3 → Al2O3 + 2Cr

Methanol (CH₃OH) is converted to bromomethane (CH₃Br) as follows: CH₃OH + HBr → CH₃Br + H₂O If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

What mass of excess reactant remains at the end of the reaction if 90.0 g of SO₂ are mixed with 100.0 g of O₂? 2SO₂ + O₂ → 2SO₃

What is the theoretical yield of aluminum that can be produced by the reaction of 60.0 g of aluminum oxide with 30.0 g of carbon according to the following chemical equation? Al₂O₃ + 3C → 2Al + 3CO

Tetraphosphorus hexaoxide is formed by the reaction of phosphorus with oxygen gas.If a mixture of 75.3 g of phosphorus and 38.7 g of oxygen produce 43.3 g of P₄O₆, what is the percent yield for the reaction?

What mass of FeS is formed if 9.42 g of Fe reacts with 68.0 g of S₈? 8Fe(s) + S₈(s) → 8FeS(s)

What mass of Cr can be produced by the reaction of 44.1 g of Cr₂O₃ with 35.0 g of Al according to the following chemical equation? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr

If 119.3 g of PCl₅ are formed from the reaction of 61.3 g Cl₂ with excess PCl₃, what is the percent yield? PCl₃(g) + Cl₂(g) → PCl₅(g)

What is the maximum number of grams of ammonia, NH₃, which can be obtained from the reaction of 10.0 g of H₂ and 80.0 g of N₂? N₂ + 3H₂ → 2NH₃

Determine the mass of precipitate (in grams) that forms when 425.1 mL of 0.122 M KI solution is reacted with excess Pb(NO₃) ₂ solution. 2KI(aq) + Pb(NO₃) ₂(aq) → PbI₂(s) + 2KNO₃(aq)

What is the theoretical yield of vanadium that can be produced by the reaction of 40.0 g of V₂O₅ with 40.0 g of calcium based on the following chemical equation? V₂O₅(s) + 5Ca(l) → 2V(l) + 5CaO(s)

Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor as follows: 4NH₃ + 5O₂ → 4NO + 6H₂O What is the maximum amount of water that may be produced if 40.0 g NH₃ and 50.0 g O₂ are mixed and allowed to react?

The first step in the Ostwald process for producing nitric acid is as follows: 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g) . If the reaction of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?

What is the theoretical yield of chromium that can be produced by the reaction of 40.0 g of Cr₂O₃ with 8.00 g of aluminum according to the chemical equation below? 2Al + Cr₂O₃ → Al₂O₃ + 2Cr