Let's start by writing the balanced reaction equation:
2H2 + O2 "\\rightarrow" 2H2O
As one can see, when there is enough oxygen, 2 molecules of hydrogen give 2 molecules of water. Therefore, when there is enough oxygen, the number of the moles of hydrogen reacted equals the number of the moles of water produced:
"n \\text{ (H}_2\\text{)} = n \\text{ (H}_2\\text{O)}"
Let's calculate the number of the moles of hydrogen in 8.00 grams of hydrogen (the number of the moles is mass "m" divided by molar mass "M", "M"(H2) =2.01568 g/mol):
"n \\text{ (H}_2\\text{)}= \\frac{m}{M} = \\frac{8.00}{2.01568} = 3.97" mol
"n \\text{ (H}_2\\text{O)} = 3.97" mol
The molar mass of water is 18.01528 g/mol. The mass of water produced is:
"m = n\\cdot M = 3.97 \\cdot 18.01528 = 71.5" g
Answer: The total mass in grams of water formed when 8.00 grams of hydrogen reacts completely with excess oxygen is 71.5 g.
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