Question 1

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 2

Aluminum metal reacts with chlorine gas to form solid aluminum chloride. What mass of chlorine gas is required to react completely with 163 g of aluminum?

Accepted Answer

The answer of Aluminum metal reacts with chlorine gas to...

Question 3

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

Manganese(II) chloride is the limiting reactant due to its stoichiometry in the reaction, leading to the production of 36.3 g of hydrochloric acid.

Question 4

Lead(II) sulfide was once used in glazing earthenware. It will also react with hydrogen peroxide to form lead(II) sulfate and water. How many grams of hydrogen peroxide are needed to react completely with 265 g of lead(II) sulfide?

Accepted Answer

The reaction between lead(II) sulfide (PbS) and hydrogen peroxide (H2O2) to form lead(II) sulfate (PbSO4) and water (H2O) can be represented by the balanced chemical equation: PbS + 4 H2O2 → PbSO4 + 4 H2O. To find the mass of H2O2 needed, first, calculate the moles of PbS using its molar mass (239.26 g/mol). For 265 g of PbS: moles of PbS = 265 g / 239.26 g/mol = 1.107 moles. From the equation, 1 mole of PbS reacts with 4 moles of H2O2. So, 1.107 moles of PbS will need 4 * 1.107 = 4.428 moles of H2O2. The molar mass of H2O2 is 34.0147 g/mol, so the mass of H2O2 needed is 4.428 moles * 34.0147 g/mol = 150.6 g, which rounds to 151 g.

Question 5

Phosphine, an extremely poisonous and highly reactive gas, reacts with oxygen gas to form tetraphosphorus decaoxide and water. PH₃(g) + O₂(g) → P₄O₁₀(s) + H₂O(g) [unbalanced] Calculate the mass of P₄O₁₀(s) formed when 225 g of PH₃ reacts with excess oxygen.

Accepted Answer

First, balance the chemical equation: 4 PH₃ + 5 O₂ → P₄O₁₀ + 6 H₂O. Calculate moles of PH₃: 225 g / 34.0 g/mol = 6.618 moles. From the balanced equation, 4 moles of PH₃ produce 1 mole of P₄O₁₀. Thus, 6.618 moles of PH₃ produce 6.618 / 4 = 1.6545 moles of P₄O₁₀. The molar mass of P₄O₁₀ is 283.9 g/mol, so the mass of P₄O₁₀ is 1.6545 moles * 283.9 g/mol = 470 g.

Question 6

What mass of sodium fluoride (used in water fluoridation and the manufacture of insecticides) is required to form 485 g of sulfur tetrafluoride? 3SCl₂(l) + 4NaF(s) → SF₄(g) + S₂Cl₂(l) + 4NaCl(s)

Accepted Answer

From the balanced chemical equation, we can see that the molar ratio of sulfur tetrafluoride to sodium fluoride is 1:4. This means that for every 1 mole of sulfur tetrafluoride produced, we need 4 moles of sodium fluoride.

To calculate the mass of sodium fluoride required to form 485 g of sulfur tetrafluoride, we first need to calculate the number of moles of sulfur tetrafluoride:

molar mass of SF4 = 108 g/mol 
moles of SF4 = mass/molar mass = 485 g / 108 g/mol = 4.49 mol

Since the molar ratio of SF4 to NaF is 1:4, we need 4 times as many moles of NaF:

moles of NaF = 4.49 mol x 4 = 17.96 mol

Finally, we can calculate the mass of NaF required:

molar mass of NaF = 42 g/mol 
mass of NaF = moles of NaF x molar mass = 17.96 mol x 42 g/mol = 754 g

Therefore, the answer is C) 754 g.

Question 7

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 8

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 9

How many grams of water could be made from 5.0 mol H₂ and 3.0 mol O₂?

Accepted Answer

The balanced chemical equation for the reaction between HS (hydrogen sulfide) and O (oxygen) to form water (H2O) and sulfur dioxide (SO2) is: 2HS + 3O -> 2H2O + 2SO2. From the given amounts, hydrogen sulfide is the limiting reactant. 5.0 mol of HS will produce 5.0 mol of H2O (since the ratio is 1:1 from the balanced equation). The molar mass of water (H2O) is approximately 18 g/mol, so 5.0 mol of H2O corresponds to 5.0 mol * 18 g/mol = 90 g of water.

Question 10

How many grams of Cl₂ can be prepared from the reaction of 16.0 g of MnO₂ and 30.0 g of HCl according to the following balanced chemical equation? MnO₂ + 4HCl → MnCl₂ + Cl₂ + 2H₂O

Accepted Answer

The answer of How many grams of Cl₂ can be...

Question 11

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 balanced chemical equation is:2Ca + O2 → 2CaOMoles of Ca = 33.7 g / 40.08 g/mol = 0.841 molMoles of O2 = 13.9 g / 32.00 g/mol = 0.434 molThe mole ratio of Ca to O2 is 2:1, so the limiting reactant is O2.Moles of CaO produced = 0.434 mol O2 × (2 mol CaO / 1 mol O2) = 0.868 molMass of CaO produced = 0.868 mol × 56.08 g/mol = 48.7 g

Question 12

Hydrochloric acid can be prepared by the following reaction: 2NaCl(s) + H₂SO₄(aq) → 2HCl(g) + Na₂SO₄(s) What mass of HCl can be prepared from 2.00 mol H₂SO₄ and 150. g NaCl?

Accepted Answer

The answer of Hydrochloric acid can be prepared by the...

Question 13

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

The answer of What mass of FeS is formed if...

Question 14

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 15

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 16

Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO₃(s) → KCl(s) + O₂(g) [unbalanced] What mass of oxygen gas is formed if 26.4 g of potassium chlorate decomposes completely?

Accepted Answer

The answer of Potassium chlorate (used in fireworks, flares, and...

Question 17

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

The answer of If 119.3 g of PCl₅ are formed...

Question 18

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

The answer of Magnesium (used in the manufacture of light...

Question 19

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

The limiting reactant is SO₂. 90.0 g of SO₂ is 1.406 moles, and 100.0 g of O₂ is 3.125 moles. The reaction requires 1 mole of O₂ for every 2 moles of SO₂, so only 0.703 moles of O₂ are needed. Therefore, 3.125 - 0.703 = 2.422 moles of O₂ remain, which is 77.5 g.

Question 20

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)

Quiz 3: Stoichiometry: Ratios of Combination

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.

Aluminum metal reacts with chlorine gas to form solid aluminum chloride. What mass of chlorine gas is required to react completely with 163 g of aluminum?

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?

Lead(II) sulfide was once used in glazing earthenware. It will also react with hydrogen peroxide to form lead(II) sulfate and water. How many grams of hydrogen peroxide are needed to react completely with 265 g of lead(II) sulfide?

Phosphine, an extremely poisonous and highly reactive gas, reacts with oxygen gas to form tetraphosphorus decaoxide and water. PH3(g) + O2(g) → P4O10(s) + H2O(g) [unbalanced] Calculate the mass of P4O10(s) formed when 225 g of PH3 reacts with excess oxygen.

What mass of sodium fluoride (used in water fluoridation and the manufacture of insecticides) is required to form 485 g of sulfur tetrafluoride? 3SCl2(l) + 4NaF(s) → SF4(g) + S2Cl2(l) + 4NaCl(s)

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?

How many grams of water could be made from 5.0 mol H2 and 3.0 mol O2?

How many grams of Cl2 can be prepared from the reaction of 16.0 g of MnO2 and 30.0 g of HCl according to the following balanced chemical equation? MnO2 + 4HCl → MnCl2 + Cl2 + 2H2O

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?

Hydrochloric acid can be prepared by the following reaction: 2NaCl(s) + H2SO4(aq) → 2HCl(g) + Na2SO4(s) What mass of HCl can be prepared from 2.00 mol H2SO4 and 150. g NaCl?

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 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

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?

Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → KCl(s) + O2(g) [unbalanced] What mass of oxygen gas is formed if 26.4 g of potassium chlorate decomposes completely?

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 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 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