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2 NH3 + 2 O2 → N2O + 3 H2O If 80.0 grams of O2 are reacted in the above reaction, how many grams of N2O will be produced (MM O2=32 g/mol; NH3=17.04 g.mol, N2O=44.02 g/mol, H2O=18.02 g/mol)?
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Is it 29.1 g?
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The answer is 55.03 grams. This is a stoichiometry problem. In stoichiometry you convert Grams of A to Mols of A to Mols of B to Grams of B. You convert from grams to mols using formula weights and you use the balanced equation for the molA-molB conversion ratio. In this case the ratio was 1/2. \[80 g O2 \times (1 mol/32grams) \times (1 mol N20/2mol O2) \times (44.02 grams/1 mol)= 55.03 grams\]
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