If you burn 46.4 g of hydrogen and produce 414 g of water, how much oxygen reacted?

1) Write the chemical reaction to state the molar relations of reactants and products:

2H2(g) + O2 (g) -> 2H2O (g)

2) Molar ratios

2 mol H2 : 1 mol O2 : 2 mol H2O

3) Convert the 46.4 g of H2 and the 414 g of H2O to moles

# moles = grams / molar mass

#moles H2 = 46.4 g / 2 g/mol = 23.2 mol

# moles H2O = 414 g / 18 g/mol = 23 mol

Then the ratio of H2 to H2O = 1.02, which means that there was plenty for the reaction, becasue the theoretical ratio (from the chemical reaction) is 1.0

The number of moles of O2 that reacts to produce  23 mol of H2O is half, becasue the theoretical ratio is 2 mol H2O : 1 mol O2

Then, 11.5 mol of oxygen reacted and that is 11.5 mol * 32.0 g/mol = 368 grams of O2.

368 g 

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