Question

Some rocket engines use a mixture of Hydrazine, $N_2{H}_4$ and Hydrogen peroxide, $H_2{O}_2$ as the propellant. The reaction is given by the following equation.

$N_2{H}_4\left(l\right)+2H_2{O}_2\left(l\right)\to N_2\left(g\right)+4H_{2}O\left(g\right)$

How much of the excess reactant remains unchanged when $0.705mol$ of $N_2{H}_4$ is mixed with $17g$ of $H_2{O}_2$?

• A. $14.6g{N}_2{H}_4$
• B. $0.25mol{H}_2{O}_2$
• C. $0.25mol{N}_2{H}_4$
• D. $8.5g{H}_2{O}_2$

Answer 1

The correct option is A $14.6g{N}_2{H}_4$

$N_2{H}_4+2H_2{O}_2⟶N_2+4H_{2}O$

Moles:

$n_{N_2{H}_4}=0.705$ $moles$

$n_{H_{2}O}=\frac{17}{34}=0.5$ $moles$

$1$ $moles$ $N_2{H}_4$ required $2$ $moles$ $H_2{O}_2$

$\therefore 0.705$ $moles$ $N_2{H}_4$ requires $1.41$ $moles$ $H_2{O}_2$

But we have only $0.5$ $moles$ $H_2{O}_2$

$\therefore N_2{H}_4$ will be leftover and it is our excess reagent.

$2$ $mole$ $H_2{O}_2$ use $1$ $mole$ $N_2{H}_4$

$\therefore 0.5$ $moles$ $H_2{O}_2$ will need $x$ $moles$ $N_2{H}_4$

$x=\frac{1\times 0.5}{2}=0.25$ $moles$

$\therefore N_2{H}_4$ left unused $=0.705-0.25=0.455$ $moles$

Mass of $N_2{H}_4$ left unreacted $=0.455\times 32=14.6$ $g$