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1. In the reaction N₂ + 3 H₂ -> 2NH3 What amount of NH3

that can be made with 1.2 mols N₂ and 2.0 mol H₂? What is the
limiting reactant ?What reactant is in excess?

2. For this reaction: C2H4 + 3 O2 ⇒ 2 CO2 + 2 H2O

a. If you start with 45g of ethylene (C₂H4),this is the limiting reagent,
how many grams of carbon dioxide (CO₂) will be produced.
b. If the actual yield is 139g CO₂, compute for the % yield.

3.
K3PO4 + Al(NO3)33KNO3 + AIPO4


a. What is the mass of potassium nitrate, ( KNO3) that is produced
when 53g of potassium phosphate, (K3PO4) react in excess of
aluminum nitrate, (Al(NO3)3
b. Compute for the % yield if the actual yield is 68g KNO3.​

User RagnaRock
by
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1 Answer

3 votes

Answer:

1. To determine the amount of NH3 that can be made with 1.2 mols of N₂ and 2.0 mol of H₂, we first need to identify the limiting reactant. We can do this by calculating the number of moles of NH3 that can be produced from each reactant:

From N₂: 1.2 mol N₂ × (2 mol NH3/1 mol N₂) = 2.4 mol NH3

From H₂: 2.0 mol H₂ × (2 mol NH3/3 mol H₂) = 1.33 mol NH3

Since the amount of NH3 produced from H₂ is less than that produced from N₂, H₂ is the limiting reactant. Therefore, the maximum amount of NH3 that can be produced is 1.33 mol NH3.

2.

a. The balanced equation for the reaction is:

C2H4 + 3 O2 ⇒ 2 CO2 + 2 H2O

The molar mass of C₂H₄ is 28 g/mol. To determine the amount of C₂H₄ in 45 g, we divide the mass by the molar mass:

45 g C₂H₄ ÷ 28 g/mol C₂H₄ = 1.61 mol C₂H₄

According to the balanced equation, 1 mole of C₂H₄ produces 2 moles of CO₂. Therefore, 1.61 moles of C₂H₄ will produce:

1.61 mol C₂H₄ × (2 mol CO₂/1 mol C₂H₄) = 3.22 mol CO₂

To convert this to grams, we multiply by the molar mass of CO₂:

3.22 mol CO₂ × 44 g/mol CO₂ = 141.68 g CO₂

Therefore, if 45 g of C₂H₄ is the limiting reagent, 141.68 g of CO₂ can be produced.

b. The theoretical yield of CO₂ is 141.68 g, but the actual yield is 139 g. To calculate the percent yield, we use the formula:

% yield = (actual yield / theoretical yield) × 100%

% yield = (139 g / 141.68 g) × 100% = 98.1%

Therefore, the percent yield is 98.1%.

3.

a. To determine the mass of KNO₃ produced when 53 g of K₃PO₄ reacts in excess of Al(NO₃)₃, we need to first balance the equation:

2 K₃PO₄ + 3 Al(NO₃)₃ → 6 KNO₃ + Al₂(PO₄)₃

The molar mass of K₃PO₄ is 212.27 g/mol. To determine the amount of K₃PO₄ in 53 g, we divide the mass by the molar mass:

53 g K₃PO₄ ÷ 212.27 g/mol K₃PO₄ = 0.25 mol K₃PO₄

According to the balanced equation, 2 moles of K₃PO₄ produce 6 moles of KNO₃. Therefore, 0.25 moles of K₃PO₄ will produce:

0.25 mol K₃PO₄ × (6 mol KNO₃/2 mol K₃PO₄) = 0.75 mol KNO₃

To convert this to grams, we multiply by the molar

User Cactustictacs
by
8.7k points
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