Answer:
The final mass of ammonia is 1.57 grams
The final mass of water is 0 grams
The final mass of magnesium nitride is 5.34 grams
Explanation:
Step 1: Data given
Mass of Mg3N2 = 10.0 grams
Molar mass of Mg3N2 = 100.95 g/mol
Mass of H2O = 5.00 grams
Molar mass H2O = 18.02 g/mol
Step 2: The balanced equation
Mg3N2 + 6H2O → 3Mg(OH)2 + 2NH3
Step 3: Calculate moles Mg3N2
Moles Mg3N2 = mass / molar mass
Moles Mg3N2 = 10.0 grams / 100.95 g/mol
Moles Mg3N2 = 0.0991 moles
Step 4: Calculate moles H20
Moles H2O = 5.00 grams / 18.02 g/mol
Moles H2O = 0.277 moles
Step 5: Calculate the limiting reactant
H2O is the limiting reactant. It will completely be consumed (0.277 moles).
Mg3N2 is in excess. There will react 0.277/6 = 0.0462 moles
There will remain 0.0991 - 0.0462 = 0.0529 moles
This is 0.0529 moles * 100.95 g/mol = 5.34 grams
Step 6: Calculate moles of NH3
For 1 mol Mg3N2 we need 6 moles H2O to produce 3 moles Mg(OH)2 and 2 moles NH3
For 0.277 moles H2O we'll have 0.277/3 = 0.0923 moles NH3
Step 7: Calculate mass NH3
Mass NH3 = moles NH3 * molar mass NH3
Mass NH3 = 0.0923 moles * 17.03 g/mol
Mass NH3 = 1.57 grams
