Answer:
34.21g of ammonia
Explanation:
N2 + 3 H2 ------->2 NH3
First find the limiting reactant
Using hydrogen
10.62 g of H2× 1.0 mole of H2/2g H2 ×2.0 mol of NH3/3.0 mol of H2× 17.0 g of NH3/1.0 mol of NH3= 60.18g of NH3
Using nitrogen
28.17 g of N2× 1.0 mole of N2/28g N2 ×2.0 mol of NH3/1.0 mol of N2 × 17.0 g of NH3/1.0 mol of NH3= 17.1g of NH3
Hence nitrogen gas is the limiting reactant.
From the reaction equation:
28.0 g of nitrogen gas gives 34.0g of ammonia
28.17 g of nitrogen gas gives 28.17×34.0/28.0= 34.21g of ammonia
Answer:
34.26 grams of NH3 will be formed
Explanation:
Step 1: Data given
Mass of nitrogen gas (N2) = 28.17 grams
Molar mass of N2 = 28.00 g/mol
Mass of hydrogen gas (H2) = 10.62 grams
Molar mass of H2 = 2.02 g/mol
Step 2: The balanced equation
N2(g) + 3H2(g) → 2NH3(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles N2 = 28.17 grams / 28.00 g/mol
Moles N2 = 1.006 moles
Moles H2 =10.62 grams / 2.02 g/mol
Moles H2 = 5.27 moles
Step 4: Calculate the limiting reactant
For 1 mon N2 we need 3 moles H2 to produce 2 moles NH3
N2 is the limiting reactant. It will completely be consumed (1.006 moles). H2 is in excess. There will react 3*1.006 = 3.018 moles
There will remain 5.27 - 3.018 = 2.252 moles
Step 5: Calculate moles NH3
For 1 mon N2 we need 3 moles H2 to produce 2 moles NH3
For 1.006 moles N2 we'll have 2*1.006 = 2.012 moles NH3
Step 6: Calculate mass NH3
Mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 2.012 moles * 17.03 g/mol
Mass NH3 = 34.26 grams
38.13 grams of NH3 will be formed
