Consider the reaction of 33.5g of hydrazine with 30.3g of hydrogen peroxide:

Consider the reaction of 33.5g of hydrazine (N2H4) with 30.3g of hydrogen peroxide (H2O2).

Here's the equation:

N2H4 + 2H2O2 ----------> N2 + 4H2O




Questions:


A. How many grams of H2O maybe produced?

B. If 29.0 grams of water are obtained, what is the percent yield?

C. How many grams of the reactant in excess would remain after the reaction in part "A" is complete?

Hello

moles N2H4 = 33.5/32 = 1.0468
moles H2O2 = 30.3/34 = 0.891
moles N2H4 reacting with 0.891 H2O2 = 0.891/2 = 0.4455
moles water produced = 2*0.891 = 1.782
g water produced = 1.782*18 = 32.08 g
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29 g water would be 29*100/38.08% = 76.15% yield
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moles N2H4 remaining = 1.0468-0.4455 = 0.6013

Regards

Hi Elegie,

Yes, I meant 32.08 g, sorry

Thanks for BA,

Ossi

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