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