Answer:
There will be produced 18.02 grams of H2O
Explanation:
Step 1: Data given
Mass of H2 = 4.0 grams
Mass of O2 = 16 grams
Molar mass of O2 = 32 g/mol
Molar mass of H2 = 2.02 g/mol
Molar mass of H2O = 18.02 g/mol
Step 2: The balanced equation
O2 + 2H2 → 2H2O
Step 3: Calculate moles of O2
Moles O2 = mass O2 / molar mass O2
Moles O2 = 16.0 grams / 32. g/mol
Moles O2 = 0.500 moles
Step 4: Calculate moles H2
Moles H2 = 4.0 grams / 2.02 g/mol
Moles H2 = 1.98 moles
Step 5: Calculate limting reactant
For 1 mol of O2 we need 2 moles of H2 to produce 2 moles of H2O
O2 is the limiting reactant. It will completely be consumed. (0.500 moles).
H2 is in excess. There will be react 2*0.500 = 1.00 moles
There will remain 1.98 - 1.00 = 0.98 moles
Step 6: Calculate moles of H2O
For 1 moles of O2 we need 2 moles of H2O
For 0.500 moles of O2 consumed, we'll produce 2*0.500 = 1.00 moles of H2O
Step 7: Calculate mass of H2O
Mass H2O = moles H2O * molar mass H2O
Mass H2O = 1.00 moles * 18.02 g/mol
Mass H2O = 18.02 grams
There will be produced 18.02 grams of H2O
