Answer:
44.4 grams of NO can be produced
Explanation:
Step 1: Data given
Mass of NO2 = 204 grams
Molar mass NO2 = 46.0 g/mol
Mass of H2O = 58.1 grams
Molar mass H2O = 18.02 g/mol
Step 2: The balanced equation
3NO2 + H2O→ 2HNO3 + NO
Step 3: Calculate moles NO2
Moles NO2 = 204 grams / 46.0 g/mol
Moles NO2 = 4.43 moles
Step 4: Calculate moles H2O
Moles H2O = 58.1 grams / 18.02 g/mol
Moles H2O = 3.22 moles
Step 5: Calculate limiting reactant
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
NO2 is the limiting reactant. It will completely be consumed (4.43 moles). H2O is in excess. there will react 4.43 /3 = 1.48 moles. There will remain 3.22 - 1.48 = 1.74 moles
Step 6: Calculate moles NO
For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO
For 4.43 moles NO2 we'll have 4.43/3 = 1.48 moles NO
Step 7: Calculate mass NO
Mass NO = 1.48 moles * 30.01 g/mol
Mass NO = 44.4 grams
44.4 grams of NO can be produced
