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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.
$N_{2} H_{4 (l)} + 2 H_{2} O_{2 (l)} \to N_{2 (g)} + 4 H_{2} O_{(g)}$

How much of the excess reactant, remains unchanged ? When 0.750 mol of $N_{2} H_{4}$ is mixed with 17 g of $H_{2} O_{2}$ ?

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16 g N_{2} H_{4}

0.25 m o l H_{2} O_{2}

0.25, m o l N_{2} H_{4}

8.5 g H_{2} O_{2}

answer is A.

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Detailed Solution

\underset{1 m o l e}{N_{2} H_{4 (l)}} + \underset{2 m o l e s}{2 H_{2} O_{(l)}} \to N_{2 (g)} + 4 H_{2} O_{(g)}

$\Rightarrow 1 m o l e N_{2} H_{4}$ can react with 2 moles of $H_{2} O_{2}$

$\Rightarrow 1 m o l e N_{2} H_{4} \to 2 \times 34 g o f H_{2} O_{2}$

$\frac{1}{4} m o l e s N_{2} H_{4} \to 17 g o f H_{2} O_{2}$

$\Rightarrow 0.25$ moles of $N_{2} H_{4}$ is reacted

The amount of unreacted $N_{2} H_{4} = 0.75 - 0.25 = 0.5 m o l e s N_{2} H_{4}$

$= 0.5 \times 32 = 16 g N_{2} H_{4}$

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