-convert g of H2 and of O2 each to mol H2O
steps 2 & 3:
-convert the one with lower mol H2O to grams
-the reactant that produces the lesser amount of H2O is the limiting reagent and is the mass of water can be obtained
Find the equation:
H2 + O2 --> H2O
Balance the equation:
2H2 + O2 --> 2H2O
Find out how many moles of H2 and O2 you've got:
4.0g H2 * 1mol H2/2g = 2 mol H2
16g O2 * 1mol O2/32g = .5mol O2
Which one will be used up before the other (aka which one is the limiting reactant?)? You can use stoichiometry to find out:
2 mol H2 = 2 mol H2/1 mol O2
1 mol of O2 is required to use up all the H2 you have. You only have .5 mol of O2. Therefore, O2 is the limiting reactant. Now you can use stoichiometry to find how much water will be produced:
.5mol O2 = 1mol O2/2mol H2O
1 mol of H2O will be produced. Now convert it to grams:
1 mol H2O * 18g/mol = 18g H2O
This could have been reasoned through without all that extra work, but that's a general method to use when the numbers become less simple. Also, I wonder if there's any better methods of formatting as I'm sure that looks confusing.
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