Question #205207

The energy from the combustion of Hydrazine, N2H4, is used to power rockets into space in the reaction:


N2H4(g)+O2(g)=N2(g)+2H2O(l) delta H degree= -627.6KJ


How many killograms of hydrazine would be necessary to produce 1.0 X 10^8 KJ of energy?


Hint: One mole of N2H4 produces 627.6KJ of energy. How many moles (and then grams) are required to produce 1.0 X 10^8 KJ of energy.


1
Expert's answer
2021-06-13T11:30:01-0400

n=1.0×108627.6=1.5933×105  moln=\frac{1.0 \times 10^8}{627.6} = 1.5933 \times 10^5 \;mol

MM(N2H4)= 32.04 g/mol

m=1.5933×105×32.04=51.05×105  g=51.05×102  kg=5105  kgm = 1.5933 \times 10^5 \times 32.04 = 51.05 \times 10^5 \;g = 51.05 \times 10^2 \;kg \\ = 5105 \;kg

Answer: 5105 kg


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