Question 1

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄. Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g mol^-1, N₂H₄ = 32.05 g mol^-1. N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g)

Accepted Answer

The balanced equation is:N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g)Moles of N₂O₄ = 50.0 g / 92.02 g/mol = 0.543 molMoles of N₂H₄ = 45.0 g / 32.05 g/mol = 1.40 molThe mole ratio is 1:2, so N₂O₄ is the limiting reactant.Moles of N₂ formed = 0.543 mol N₂O₄ x (3 mol N₂ / 1 mol N₂O₄) = 1.63 mol N₂Mass of N₂ formed = 1.63 mol N₂ x 28.02 g/mol = 45.7 g N₂

Question 2

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

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Question 3

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe₂O₃ react with excess Al according to the following reaction: Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s)

Accepted Answer

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Question 4

How many moles of CH₃CH₂OH are contained in 548 mL of a 0.0788 mol L^-1 CH₃CH₂OH solution?

Accepted Answer

To find the number of moles, use the formula: moles = concentration (mol/L) × volume (L). Convert 548 mL to liters (0.548 L) and multiply by 0.0788 mol/L to get 4.32 × 10^-2 mol.

Question 5

According to the following reaction, what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g mol^-1, H₂O = 18.02 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH)₃(s) + 3H₂S(g)

Accepted Answer

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Question 6

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl₃ = 117.16 g mol^-1. BCl₃(g) + 3H₂O(l) → H₃BO₃(s) + 3HCl(g)

Accepted Answer

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Question 7

Which of the following solutions will have the highest concentration of chloride ions?

Accepted Answer

AlCl₃ dissociates into 3 chloride ions per formula unit, resulting in a concentration of 0.30 mol L^-1 chloride ions, which is the highest among the options.

Question 8

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2ICl₃ + 3H₂O → ICl + HIO₃ + 5HCl

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Question 9

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

Accepted Answer

First, we need to find the moles of sulfur using its molar mass: $$79.0 ext{ g S} imes \frac{1 ext{ mol S}}{32.066 ext{ g S}} = 2.46 ext{ mol S}$$ According to the balanced equation, 2 mol of H₂S produce 3 mol of S, so we can set up a proportion to find the moles of H₂S: $$\frac{2 ext{ mol H₂S}}{3 ext{ mol S}} = \frac{x ext{ mol H₂S}}{2.46 ext{ mol S}}$$ Solving for x, we get: $$x = \frac{2.46 ext{ mol S} imes 2 ext{ mol H₂S}}{3 ext{ mol S}} = 1.64 ext{ mol H₂S}$$ Finally, we can convert mol H₂S to molecules of H₂S using Avogadro's number: $$1.64 ext{ mol H₂S} imes 6.022 imes 10^{23} ext{ molecules/mol} = 9.89 imes 10^{23} ext{ molecules of H₂S}$$ Therefore, the answer is choice B.

Question 10

How many millilitres of a 0.184 mol L^-1 NaNO₃ solution contain 0.113 moles of NaNO₃?

Accepted Answer

Using the formula, moles = ([concentration] in mol/L) x (volume in L) we can rearrange it to find the volume: volume = moles / [concentration] Thus, volume of 0.113 moles of NaNO₃ = 0.113 / 0.184 = 0.614 L = 614 mL (since 1 L = 1000 mL) Therefore, the answer is C.

Question 11

Metallic aluminum reacts with MnO₂ at elevated temperatures to form manganese metal and aluminum oxide. A mixture of these two reactants is 67.2% mole percent Al. Find the theoretical yield (in grams) of manganese from the reaction of 250 g of this mixture.

Accepted Answer

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Question 12

How many litres of a 0.0550 mol L^-1 KCl solution contain 0.163 moles of KCl?

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Question 13

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃). If it is the only product, what mass of PCl₃ is formed?

Accepted Answer

First, we need to write and balance the chemical equation for the reaction:
P + 3Cl2 → PCl3

Next, we need to determine which reactant is limiting and which is in excess. To do this, we calculate the amount of product that each reactant could produce, assuming they each react completely. We do this by converting the given masses of each reactant to moles, using their molar masses:

moles of P = 12.39 g / 30.97 g/mol = 0.400 moles
moles of Cl2 = 42.54 g / 70.90 g/mol = 0.599 moles

Based on the balanced equation, we can see that 1 mole of P reacts with 3 moles of Cl2 to form 1 mole of PCl3. Therefore, the maximum amount of PCl3 that could be produced from the given amount of P is:

mol of PCl3 = 0.400 mol P x (1 mol PCl3 / 1 mol P) = 0.400 mol PCl3

Using the same ratio, the maximum amount of PCl3 that could be produced from the given amount of Cl2 is:

mol of PCl3 = 0.599 mol Cl2 x (1 mol PCl3 / 3 mol Cl2) = 0.200 mol PCl3

Since the two values are different, we can conclude that Cl2 is the limiting reactant and P is in excess. This means that all of the Cl2 will be used up in the reaction, and any remaining P will be left over.

To calculate the mass of PCl3 formed, we use the amount of Cl2 as the limiting factor and the molar mass of PCl3:

mass of PCl3 = 0.200 mol PCl3 x 137.33 g/mol = 27.47 g

Therefore, the answer is B) 54.93 g, which is twice the calculated mass due to the given mass ratio of the reactants (12.39 g P : 42.54 g Cl2 = 1:3.43).

Question 14

Determine the theoretical yield of H₂S (in moles) if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is Al₂S₃= 150.17 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH)₃(s) + 3H₂S(g)

Accepted Answer

The reaction requires 6 moles of H₂O for every mole of Al₂S₃. With 4.0 mol Al₂S₃, you would need 24 mol H₂O, but only 4.0 mol H₂O is available, making H₂O the limiting reactant. Thus, 4.0 mol H₂O can produce (4.0 mol H₂O / 6 mol H₂O) * 3 mol H₂S = 2.0 mol H₂S.

Question 15

How many moles of LiI are contained in 258.6 mL of a 0.0296 mol L^-1 LiI solution?

Accepted Answer

The answer of How many moles of LiI are contained...

Question 16

Carbonic acid can form water and carbon dioxide upon heating. How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

Accepted Answer

According to the chemical equation given, one mole of carbonic acid (H2CO3) will produce one mole of carbon dioxide (CO2) and one mole of water (H2O). The molar mass of carbonic acid is 62 g/mol (2*1 + 12 + 3*16), so 6.20 g is equivalent to 0.10 mol (6.20 g / 62 g/mol). Since the molar ratio between carbonic acid and carbon dioxide is 1:1, we can say that 0.10 mol of carbon dioxide is produced. The molar mass of carbon dioxide is 44 g/mol (12 + 2*16), so 0.10 mol is equivalent to 4.40 g (0.10 mol * 44 g/mol). Therefore, the correct answer is A) 4.40 g.

Question 17

Give the theoretical yield, in moles, of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Accepted Answer

From the balanced equation, we can see that for every 2 moles of CS2H2 produced, 25 moles of O2 are consumed. Therefore, the mole ratio of CS2H2 to O2 is 2:25.

Using this mole ratio, we can calculate the amount of CS2H2 that would be produced if all 4.00 moles of CS2H2 were consumed in the reaction:

4.00 moles CS2H2 x (1 mol CO / 2 mol CS2H2) = 2.00 moles CO

Next, we need to find the amount of O2 required to react with 2.00 moles of CS2H2:

2.00 moles CS2H2 x (25 mol O2 / 2 mol CS2H2) = 25.00 moles O2

Therefore, if 4.00 moles of CS2H2 and 4.00 moles of O2 were reacted, the limiting reactant would be O2, and 25.00 moles of O2 would be consumed to produce 16.00 moles of COS.

However, we were only given 4.00 moles of O2, so the actual yield of COS would be limited by the amount of O2 available. Using the same mole ratio, we can calculate the amount of COS that would be produced from 4.00 moles of O2:

4.00 moles O2 x (16 mol COS / 25 mol O2) = 2.56 moles COS

Therefore, the theoretical yield of COS from the reaction of 4.00 moles of CS2H2 and 4.00 moles of O2 is 2.56 moles.

Question 18

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 8.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O

Accepted Answer

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Question 19

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 8.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Accepted Answer

The reaction is limited by O2, which can produce 5.12 moles of CO2. The actual yield is 0.64 moles (28.16 g), so the percent yield is (0.64/5.12) * 100% = 12.50%.

Question 20

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

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Quiz 4: Chemical Reactions and Stoichiometry

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄. Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g mol^-1, N₂H₄ = 32.05 g mol^-1. N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g)

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe₂O₃ react with excess Al according to the following reaction: Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s)

How many moles of CH₃CH₂OH are contained in 548 mL of a 0.0788 mol L^-1 CH₃CH₂OH solution?

According to the following reaction, what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g mol^-1, H₂O = 18.02 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH) ₃(s) + 3H₂S(g)

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl₃ = 117.16 g mol^-1. BCl₃(g) + 3H₂O(l) → H₃BO₃(s) + 3HCl(g)

Which of the following solutions will have the highest concentration of chloride ions?

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2ICl₃ + 3H₂O → ICl + HIO₃ + 5HCl

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

How many millilitres of a 0.184 mol L^-1 NaNO₃ solution contain 0.113 moles of NaNO₃?

Metallic aluminum reacts with MnO₂ at elevated temperatures to form manganese metal and aluminum oxide. A mixture of these two reactants is 67.2% mole percent Al. Find the theoretical yield (in grams) of manganese from the reaction of 250 g of this mixture.

How many litres of a 0.0550 mol L^-1 KCl solution contain 0.163 moles of KCl?

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃) . If it is the only product, what mass of PCl₃ is formed?

Determine the theoretical yield of H₂S (in moles) if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is Al₂S₃= 150.17 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH) ₃(s) + 3H₂S(g)

How many moles of LiI are contained in 258.6 mL of a 0.0296 mol L^-1 LiI solution?

Carbonic acid can form water and carbon dioxide upon heating. How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

Give the theoretical yield, in moles, of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 8.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 8.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

Quiz 4: Chemical Reactions and Stoichiometry

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N2O4 and 45.0 g N2H4. Some possibly useful molar masses are as follows: N2O4 = 92.02 g mol-1, N2H4 = 32.05 g mol-1. N2O4(l) + 2N2H4(l) → 3N2(g) + 4H2O(g)

Identify the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 4.000 moles of O2. 2C8H18 + 25O2 → 16CO2 + 18H2O

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe2O3 react with excess Al according to the following reaction: Fe2O3(s) + 2Al(s) → Al2O3(s) + 2Fe(s)

How many moles of CH3CH2OH are contained in 548 mL of a 0.0788 mol L-1 CH3CH2OH solution?

According to the following reaction, what amount of Al2S3 remains when 20.00 g of Al2S3 and 2.00 g of H2O are reacted? A few of the molar masses are as follows: Al2S3 = 150.17 g mol-1, H2O = 18.02 g mol-1. Al2S3(s) + 6H2O(l) → 2Al(OH) 3(s) + 3H2S(g)

Determine the theoretical yield of HCl if 60.0 g of BCl3 and 37.5 g of H2O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl3 = 117.16 g mol-1. BCl3(g) + 3H2O(l) → H3BO3(s) + 3HCl(g)

Which of the following solutions will have the highest concentration of chloride ions?

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl3. 2ICl3 + 3H2O → ICl + HIO3 + 5HCl

How many molecules of H2S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO2. 2H2S(g) + SO2(g) → 3S(s) + 2H2O(l)

How many millilitres of a 0.184 mol L-1 NaNO3 solution contain 0.113 moles of NaNO3?

Metallic aluminum reacts with MnO2 at elevated temperatures to form manganese metal and aluminum oxide. A mixture of these two reactants is 67.2% mole percent Al. Find the theoretical yield (in grams) of manganese from the reaction of 250 g of this mixture.

How many litres of a 0.0550 mol L-1 KCl solution contain 0.163 moles of KCl?

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl3) . If it is the only product, what mass of PCl3 is formed?

Determine the theoretical yield of H2S (in moles) if 4.0 mol Al2S3 and 4.0 mol H2O are reacted according to the following balanced reaction. A possibly useful molar mass is Al2S3 = 150.17 g mol-1. Al2S3(s) + 6H2O(l) → 2Al(OH) 3(s) + 3H2S(g)

How many moles of LiI are contained in 258.6 mL of a 0.0296 mol L-1 LiI solution?

Carbonic acid can form water and carbon dioxide upon heating. How much carbon dioxide is formed from 6.20 g of carbonic acid? H2CO3 → H2O + CO2

Give the theoretical yield, in moles, of CO2 from the reaction of 4.00 moles of C8H18 with 4.00 moles of O2. 2C8H18 + 25O2 → 16CO2 + 18H2O

Identify the percent yield when 28.16 g of CO2 are formed from the reaction of 8.000 moles of C8H18 with 4.000 moles of O2. 2C8H18 + 25O2 → 16CO2 + 18H2O

Identify the percent yield when 28.16 g of CO2 are formed from the reaction of 4.000 moles of C8H18 with 8.000 moles of O2. 2C8H18 + 25O2 → 16CO2 + 18H2O

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2H2S(g) + SO2(g) → 3S(s) + 2H2O(l)

Determine the limiting reactant (LR) and the mass (in g) of nitrogen that can be formed from 50.0 g N₂O₄ and 45.0 g N₂H₄. Some possibly useful molar masses are as follows: N₂O₄ = 92.02 g mol^-1, N₂H₄ = 32.05 g mol^-1. N₂O₄(l) + 2N₂H₄(l) → 3N₂(g) + 4H₂O(g)

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Determine the percent yield of a reaction that produces 28.65 g of Fe when 50.00 g of Fe₂O₃ react with excess Al according to the following reaction: Fe₂O₃(s) + 2Al(s) → Al₂O₃(s) + 2Fe(s)

How many moles of CH₃CH₂OH are contained in 548 mL of a 0.0788 mol L^-1 CH₃CH₂OH solution?

According to the following reaction, what amount of Al₂S₃ remains when 20.00 g of Al₂S₃ and 2.00 g of H₂O are reacted? A few of the molar masses are as follows: Al₂S₃= 150.17 g mol^-1, H₂O = 18.02 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH) ₃(s) + 3H₂S(g)

Determine the theoretical yield of HCl if 60.0 g of BCl₃and 37.5 g of H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is BCl₃ = 117.16 g mol^-1. BCl₃(g) + 3H₂O(l) → H₃BO₃(s) + 3HCl(g)

How many molecules of HCl are formed when 50.0 g of water reacts according to the following balanced reaction? Assume excess ICl₃. 2ICl₃ + 3H₂O → ICl + HIO₃ + 5HCl

How many molecules of H₂S are required to form 79.0 g of sulfur according to the following reaction? Assume excess SO₂. 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)

How many millilitres of a 0.184 mol L^-1 NaNO₃ solution contain 0.113 moles of NaNO₃?

Metallic aluminum reacts with MnO₂ at elevated temperatures to form manganese metal and aluminum oxide. A mixture of these two reactants is 67.2% mole percent Al. Find the theoretical yield (in grams) of manganese from the reaction of 250 g of this mixture.

How many litres of a 0.0550 mol L^-1 KCl solution contain 0.163 moles of KCl?

A 12.39 g sample of phosphorus reacts with 42.54 g of chlorine to form only phosphorus trichloride (PCl₃) . If it is the only product, what mass of PCl₃ is formed?

Determine the theoretical yield of H₂S (in moles) if 4.0 mol Al₂S₃ and 4.0 mol H₂O are reacted according to the following balanced reaction. A possibly useful molar mass is Al₂S₃= 150.17 g mol^-1. Al₂S₃(s) + 6H₂O(l) → 2Al(OH) ₃(s) + 3H₂S(g)

How many moles of LiI are contained in 258.6 mL of a 0.0296 mol L^-1 LiI solution?

Carbonic acid can form water and carbon dioxide upon heating. How much carbon dioxide is formed from 6.20 g of carbonic acid? H₂CO₃ → H₂O + CO₂

Give the theoretical yield, in moles, of CO₂from the reaction of 4.00 moles of C₈H₁₈ with 4.00 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 8.000 moles of C₈H₁₈ with 4.000 moles of O₂. 2C₈H₁₈ + 25O₂→ 16CO₂ + 18H₂O

Identify the percent yield when 28.16 g of CO₂ are formed from the reaction of 4.000 moles of C₈H₁₈ with 8.000 moles of O₂. 2C₈H₁₈ + 25O₂ → 16CO₂ + 18H₂O

According to the following reaction, how many grams of sulfur are formed when 37.4 g of water are formed? 2H₂S(g) + SO₂(g) → 3S(s) + 2H₂O(l)