Asked by Elizabeth
Show how you would calculate the % yield of hydrogen if 40.0 grams of magnesium react with an excess of nitric acid producing 1.70 grams of hydrogen gas.
Mg + 2 HNO3 ---> Mg(NO3)2 + H2
Mg + 2 HNO3 ---> Mg(NO3)2 + H2
Answers
Answered by
DrBob222
mols Mg = grams/atomic mass = ?
Using the coefficients in the balanced equation, convert mols Mg to mols H2. You can see it's a 1:1 ratio; therefore, mols Mg = mols H2 produced.
Then convert mols H2 to grams. g H2 = mols H2 x molar mass H2 = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is listed as 1.70 g.
% yield = (AY/TY)*100 = >
Using the coefficients in the balanced equation, convert mols Mg to mols H2. You can see it's a 1:1 ratio; therefore, mols Mg = mols H2 produced.
Then convert mols H2 to grams. g H2 = mols H2 x molar mass H2 = ? This is the theoretical yield (TY). The actual yield (AY) in the problem is listed as 1.70 g.
% yield = (AY/TY)*100 = >
Answered by
Elizabeth
Thanks @DrBob222
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