One compound under investigation for use as a lightweight rocket fuel is dimethylhydrazine (60.10 g/mol) .It reacts with dinitrogen tetroxide (92.01 g/mol) according to the following reaction: (CH₃ ) ₂ N₂H₄ (l) +N₂O₄ (l) $\rightarrow$ 3 N₂ (g) + 4H₂O(g) +0 2 CO₂ (g)
If 150 g of (CH₃) ₂N₂H₄ react with excess N₂O₄ at 473 K and 760 torr, what volume of CO₂ gas will form?

Question

One compound under investigation for use as a lightweight rocket fuel is dimethylhydrazine (60.10 g/mol) .It reacts with dinitrogen tetroxide (92.01 g/mol) according to the following reaction: (CH₃ ) ₂ N₂H₄ (l) +N₂O₄ (l)

\rightarrow

3 N₂ (g) + 4H₂O(g) +0 2 CO₂ (g)
If 150 g of (CH₃) ₂N₂H₄ react with excess N₂O₄ at 473 K and 760 torr, what volume of CO₂ gas will form?

Quizplus · Accepted Answer

The reaction shows that 1 mole of dimethylhydrazine reacts to produce 2 moles of CO2 gas. Given the molar mass of dimethylhydrazine (60.10 g/mol), 150 g corresponds to 150 g / 60.10 g/mol = 2.495 moles. This will produce 2 * 2.495 = 4.99 moles of CO2. Using the ideal gas law (PV=nRT) and given conditions (T = 473 K, P = 760 torr = 1 atm), the volume V = nRT/P. With R = 0.0821 L·atm/(mol·K), V = 4.99 moles * 0.0821 L·atm/(mol·K) * 473 K / 1 atm = approximately 190 L.

One Compound Under Investigation for Use as a Lightweight Rocket →\rightarrow→

One Compound Under Investigation for Use as a Lightweight Rocket $\rightarrow$