Answer:
6.16 grams of NH3 will be produced. N2 is the limiting reactant. H2 is in excess, there will remain 4.30 grams of H2
Explanation:
Step 1: Data given
Mass of nitrogen gas(N2) = 5.07 grams
Mass of hydrogen gas (H2)
Molar mass N2 = 28.0 g/mol
Molar mass 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 = 5.07 grams / 28.0 g/mol
Moles N2 = 0.181 moles
Moles H2 = 5.40 grams / 2.02 g/mol
Moles H2 = 2.67 moles
Step 4: Calculate the 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 (0.181 moles). H2 is in excess. There will react 3*0.181 = 0.543 moles
There will remain 2.67 - 0.543 = 2.127 moles
Step 5: Calculate mass of H2 that remains
Mass H2 = 2.127 moles * 2.02 g/mol
Mass H2 = 4.30 grams
Step 6: Calculate moles ammonia
For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3
For 0.181 moles N2 we need 2*0.181 = 0.362 moles NH3
Step 7: Calculate mass NH3
Mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.362 moles * 17.01 g/mol
Mass NH3 = 6.16 grams
6.16 grams of NH3 will be produced. N2 is the limiting reactant. H2 is in excess, there will remain 4.30 grams of H2
