What is the theoretical yield of Ammonia if 14.29g of N and 10.85g of H are allo

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What is the theoretical yield of Ammonia if 14.29g of N2 and 10.85g of H2 are allowed to react? ANSWER: Step 1: Balanced equation. N2 + 3H2 ------> 2NH3 Step 2: Finding limiting reactant. Convert N2 mass to moles: 14.29 g N2 x (1 mole N2/28.02 g N2) = 0.51 moles N2. Convert N2 moles to H2 moles:0.51 moles N2 x (3 molesH2/1 mole N2) = 1.53 moles H2 Convert H2 moles to Mass : 1.53 moles H2 x (2.016 g H2 /1 mole H2) = 3.084 g H2. So 14.29 g N2 need only 3.084 g of H2. But we have 10.85 g H2. So H2 has (10.85-3.084) =7.766 g excess reactant. Limiting reactant is Nitrogen (N2). Step 3: Theoretical Yield: Convert N2 mass to moles: 14.29 g N2 x (1 mole N2/28.02 g N2) = 0.51 moles N2. Convert N2 moles to NH3 moles: 0.51 moles N2 x (2 moles NH3/1 mole N2) = 1.02 moles NH3 Convert NH3 moles to Mass: 1.02 moles NH3 x(17.03 g/1 mole NH3) = 17.37 g NH3. So the theoretical Yield of Ammonia (NH3) is 17.37 g.

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