Percent Yield

[Anjali_Kumar1F]

I am confused on how to do this I though I just divide 25.2g/35g and multiply it by 100. I got 72% but the answer is 76.6%. What are the steps to this problem?
Fundamentals M1
Hydrazine, N2H4, is an oily liquid used as a rocket fuel. It can be prepared in water by oxidizing ammonia with hypochlorite ions: 2 NH3g + ClOaq ---> N2H4aq + Cl+aq + H2Ol. When 35.0 g of ammonia reacted with an excess of hypochlorite ion, 25.2 g of hydrazine was produced. What is the percentage yield of hydrazine?

[Bingcui Guo]

I think your theoretical yield is wrong. There are 2.0588 moles of NH3 in the reactant and they would produce 2.0588 moles of N2H4 which is about 32.94g produced. In this case, 25.5/32.95*100% is 76.5%

[Ian Marquez 2K]

When you calculate a percent yield problem, make sure to convert the limiting reactant into moles. From there, using the stoichiometric coefficients, you can calculate the moles of the theoretical yield of the product (N2H4). Since the ratio of the coefficients is 2:1, there will be half as many moles of N2H4 as there is ammonium (NH3). After this, convert these moles into grams and take the actual yield of N2H4 (25.2g) and divide it by the theoretical yield that was calculated with the moles converted into grams (32.94g). Then just use the actual yield/theoretical yield multiplied by 100 formula and your answer should come out to 76.5%

[Anjali_Kumar1F]

ok that makes sense!

[Ashley Odibo Dis3E]

Can someone explain what I'm doing wrong? I keep getting 76.5 instead of 76.6. I converted the 35g of NH3 to grams of N2H4 and divided that by the actual yield and my answer continuously comes out to 76.50%

[2c_britneyly]

Your method sounds correct, and the other people who previously answered got 76.5% too. The most likely reason for the difference in percentage is that the book may have used a different molar mass for the compound N2H4.