4NH3+7O2 yields 4NO2+6H2O
If 8g of NH3 reacted with excess O2, how many g of H2O will be produced?
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To find out how many grams of H2O will be produced, you need to use stoichiometry—a branch of chemistry that deals with the relative quantities of reactants and products in chemical reactions. The balanced equation you provided is:
4NH3 + 7O2 → 4NO2 + 6H2O
From the equation, you can see that 4 moles of NH3 react to produce 6 moles of H2O.
Step 1: Convert grams of NH3 to moles
To do this, you need the molecular weight of NH3. The molecular weight of nitrogen (N) is 14.01 g/mol, and the molecular weight of hydrogen (H) is 1.008 g/mol. So the molecular weight of NH3 is:
(3 x 1.008 g/mol) + (1 x 14.01 g/mol) = 17.03 g/mol
Using the molecular weight, you can find the number of moles of NH3:
8 g NH3 × (1 mol NH3/17.03 g NH3) = 0.47 mol NH3
Step 2: Use stoichiometry to find moles of H2O
Using the balanced equation, you know that 4 moles of NH3 react to produce 6 moles of H2O. So the ratio is 6 moles of H2O/4 moles of NH3.
0.47 mol NH3 × (6 mol H2O/4 mol NH3) = 0.705 mol H2O
Step 3: Convert moles of H2O to grams
To convert moles to grams, you need the molecular weight of H2O. The molecular weight of hydrogen (H) is 1.008 g/mol, and the molecular weight of oxygen (O) is 16 g/mol. So the molecular weight of H2O is:
(2 x 1.008 g/mol) + (16 g/mol) = 18.02 g/mol
Using the molecular weight, you can find the number of grams of H2O:
0.705 mol H2O × (18.02 g H2O/1 mol H2O) = 12.71 g H2O
Therefore, if 8 grams of NH3 reacted with excess O2, approximately 12.71 grams of H2O will be produced.