Bo reacted 21.2 grams of N2H4 w/ 29.2 grams of H202 producing nitrogen gas & water.  What is the % yield of water if Bo produced 23.6 g H2O?

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bandmanjoe | Middle School Teacher | (Level 2) Senior Educator

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According to the law of conservation of mass, matter is neither created nor destroyed, it just changes forms.  What this means is the mass you end up with should equal the mass you started with.  If you add the 21.2 grams of N2H4 with the 29.2 grams of H2O2, you get a total of 50.4 grams of mass of reactants.  If Bo produces 23.6 grams of water, to get the yield of water, simply divide that by the 50.4 grams of reactant mass, and multiply by 100% to get the percentage yield of water.

23.6 g/50.4 g = .468 g x 100% = 46.8%

So according to these calculations, Bo yielded 46.8 % water from the chemical reaction.  The remainder of the mass was nitrogen gas, 26.8 grams of nitrogen gas.  This would also represent 54.2 % yield for nitrogen gas from the same reaction.  Matter is neither created nor destroyed, it just changes from one form to another.

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