Given that 4 NH3 + 5 O2 4 NO + 6 H2O, if 3.00 mol NH3 were made to react with excess of oxygen gas, the amount of H2O formed would be?
To determine the amount of H2O formed when 3.00 mol of NH3 reacts with an excess of O2, we will use stoichiometry. The balanced equation shows that 4 mol of NH3 reacts to produce 6 mol of H2O.
First, we need to find the number of moles of H2O produced by finding the mole ratio between NH3 and H2O.
Mole ratio NH3 to H2O:
1 mol NH3 : 6 mol H2O
Since we have 3.00 mol of NH3, we can set up a proportion to find the number of moles of H2O produced:
3.00 mol NH3 / 1 = x mol H2O / 6
Cross-multiplying and solving for x:
x = (3.00 mol NH3 * 6) / 1 = 18.00 mol H2O
Therefore, when 3.00 mol of NH3 reacts with an excess of O2, the amount of H2O formed would be 18.00 mol.