1. If 3.00g of ammonia reacts with 40.00g of oxygen, what is the theoretical yield of nitrogen monoxide?

2. If only 10.00g of nitrogen monoxide was isolated from the products, what was the percent yield of nitrogen monoxide?

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)