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Question

Some rocket engines use a mixture of Hydrazine, N2H4 and Hydrogen peroxide, H2O2 as the propellant. The reaction is given by the following equation.


N2H4(l)+2H2O2(l)N2(g)+4H2O(g)

How much of the excess reactant remains unchanged when 0.705 mol of N2H4 is mixed with 17 g of H2O2?

A
14.6 g N2H4
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B
0.25 mol H2O2
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C
0.25 mol N2H4
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D
8.5 g H2O2
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Solution

The correct option is A 14.6 g N2H4
N2H4+2H2O2N2+4H2O

Moles:
nN2H4=0.705 moles
nH2O=1734=0.5 moles

1 moles N2H4 required 2 moles H2O2

0.705 moles N2H4 requires 1.41 moles H2O2

But we have only 0.5 moles H2O2

N2H4 will be leftover and it is our excess reagent.

2 mole H2O2 use 1 mole N2H4

0.5 moles H2O2 will need x moles N2H4

x=1×0.52=0.25 moles

N2H4 left unused =0.7050.25=0.455 moles

Mass of N2H4 left unreacted =0.455×32=14.6 g


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