if 2.3g of H2 reacts with .5g O2, what is limiting? Produces .4g H2O what is the

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Explanation / Answer

2H2 + O2 ––––> 2H2O 100g H2 (1 mol H2 / 2g H2) = 50 mol H2 / 2 = 25 100g O2 (1 mol O2 / 32g O2) = 3.125 mol O2 The oxygen will run out first, therefore it is limiting. Now use stoichiometry to find how much water is produced with 100g of O2 100g O2 (1 mol O2 / 32g O2)(1 mol O2 / 1 mol H2O) = 3.125 mol H20 To find how much remains of the other one, use the oxygen to find how much you used, then subtract that value to find how much remains: 100g O2 ( 1mol O2 / 32g O2)(2 mol H2 / 1 mol O2) (2 g H2 / 1 mol H2) = 12.5g H2 used 100 - 12.5 = 87.5g H2 remains

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