Answer:
The theoretical yield of nitrogen monoxide = 5.28 grams
Explanation:
Step 1: Data given
Mass of ammonia = 3.00 grams
Molar mass NH3 = 17.03 g/mol
mass of oxygen = 40.00 grams
Molar mass O2 = 32.0 g/mol
Step 2: The balanced equation
4NH3 + 5O2 → 4NO + 6H2O
Step 3: Calculate moles
Moles = mass / molar mass
Moles NH3 = 3.00 grams / 17.02 g/mol
Moles NH3 = 0.176 moles
Moles O2 = 40.00 grams / 32.0 g/mol
Moles O2 = 1.25 moles
Step 4: Calculate limiting reactant
For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O
NH3 is the limiting reactant. It will completely be consumed (0.176 moles)
O2 is in excess. There will react 5/4 * 0.176 = 0.220 moles
There will remain 1.25 - 0.22 = 1.03 moles
Step 5: Calculate moles NO
For 4 moles NH3 we need 5 moles O2 to produce 4 moles NO and 6 moles H2O
For 0.176 moles NH3 we'll have 0.176 moles NO
Step 6: Calculate theoretical yield of NO
Mass NO = moles NO * molar mass NO
Mass NO = 0.176 moles * 30.01 g/mol
Mass NO = 5.28 grams
Step 7: Calculate percent yield
% yield = (10.00 grams / 5.28 grams) *100 %
% yield = 189 %
⇒ Since the theoretical yield is less than 10.0 grams. I think the data are not correct ( may be 30.00 grams of NH3)
