Answer:
The theoretical yield of NH3 produced by the reaction is 16.89 grams
Explanation:
Step 1: Data given
Mass of N2 = 13.90 grams
Molar mass of N2 = 28.0 g/mol
Mass H2 = 6.440 grams
Molar mass H2 = 2.02 g/mol
Step 2: The balanced equation
N2(g) + 3 H2(g) → 2 NH3(g)
Step 3: Calculate moles
Moles = mass / molar mass
Moles N2 = 13.90 grams / 28.0 g/mol
Moles N2 = 0.496 moles
Moles = 6.440 grams / 2.02 g/mol
Moles H2 = 3.19 moles
Step 4: Calculate limiting reactant
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
N2 is the limiting reactant. It will completely be consumed (0.496 moles)
H2 is in excess. There will react 3*0.496 = 1.488 moles
there will remain 3.19 - 1.488 = 1.702 moles
Step 5: Calculate moles NH3
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.496 moles N2 we will have 2*0.496 = 0.992 moles NH3
Step 6: Calculate mass NH3
Mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.992 moles * 17.03 g/mol
Mass NH3 = 16.89 grams
The theoretical yield of NH3 produced by the reaction is 16.89 grams
