Question

Hydrazine (N2H4), a rocket fuel , reacts with oxygen to form nitrogen gas and water vapor. The reaction is represented with the equation: N2H4(l) O2(g) → N2(g) 2H2O(g) How many grams of hydrazine(N2H4) are needed to produce 96. 0g water(H2O)?.

Answer 1

85.33 grams of hydrazine is needed for the reaction.

Rocket fuel contains the reaction between hydrazine and oxygen to break and form nitrogen and water vapour along with steam as it is an exothermic reaction.

How to calculate the hydrazine amount?

The reaction can be shown as:

$\rm N_{2}H_{4}(l) + O_{2}(g) \rightarrow N_{2}(g) + 2H_{2}O(g)$

From the periodic chart, it can be said that:

• Mass of nitrogen = 14 gm
• Mass of oxygen = 16 gm
• Mass of hydrogen = 1 gm

The molar mass of hydrazine will be:

$\begin{aligned} \rm Molar \;mass \;of \rm \;N_{2}H_{4}&= 2(14) + 4(1) \\\\&= 32 \;\rm grams\end{aligned}$

Mass of water:

$\begin{aligned}&= 2(1) + 16 \\\\&= 18 \;\rm grams\end{aligned}$

From the reaction, it can be said that,

1 mole of $\rm N_{2}H_{4}$ = 2 moles of $\rm H_{2}O$

32 grams of $\rm N_{2}H_{4}$  = ? grams of water

$\begin{aligned}\text{Gram of water} &= 18 \times 2\\\\&= 36 \;\rm gm\end{aligned}$

So, for 96 gm of water the hydrazine produced will be,

$\begin{aligned}\text{mass of }\rm N_{2}H_{4} &= \dfrac{96 \times 32}{36}\\\\&= 85.3334 \;\rm grams \end{aligned}$

Therefore, 85.33 grams of hydrazine is needed for the reaction.

Learn more about rocket fuel here:

brainly.com/question/10179270