Question

How many liters of water vapor can be produced if 108 grams of methane gas (CH4) are combusted at 312 K and 0.98 atm? Show all work. Pls help! WILL MARK AS BRAINLIEST

Answer 1

Answer:

Approximately 352 liters of water vapor will be produced.

Assumption: the water vapor here behaves like an ideal gas.

Explanation:

Relative atomic mass data from a modern periodic table:

• Carbon: 12.011;
• Hydrogen: 1.008.

Burning methane $\rm CH_4$ in excess oxygen will produce carbon dioxide and water. The balanced equation will be

$\rm CH_4\; (g) + 2\; O_2\;(g) \to CO_2\;(g) + 2\; H_2O\;(g)$.

Hint: start by assign "1" to methane while balancing this equation.

How many moles of molecules in 108 gram of methane?

Molar mass of methane:

$M(\rm CH_4) = 12.011 + 4\time 1.008 =16.043\;g\cdot mol^{-1}$.

$\displaystyle n(\rm CH_4) = \frac{m}{M} = \frac{108}{16.043} = 6.73191\;mol$.

How many moles of water molecules will be produced?

Consider the ratio between the coefficient in front of water and that in front of methane in the equation:

$\displaystyle \frac{n(\mathrm{H_2O})}{n(\mathrm{CH_4})} = 2$.

$\displaystyle n(\mathrm{H_2O}) = n(\mathrm{CH_4})\cdot \frac{n(\mathrm{H_2O})}{n(\mathrm{CH_4})} = \rm 2\times 6.73191\;mol = 13.4638\;mol$.

Assume that water vapor here behaves like an ideal gas.

Ideal gas constant in liters and $\rm atm$:

$R \rm \approx 0.0820573\;L\cdot atm \cdot K^{-1}\cdot mol^{-1}$. (NIST.)

Make sure that the temperature here is in degrees Kelvins.

$\displaystyle \begin{aligned}V &= \frac{n\cdot R\cdot T}{P} \\ &= \rm \frac{13.4638\;mol\times 0.0820573\;L\cdot atm \cdot K^{-1}\cdot mol^{-1}\times 312\; K}{0.98\; atm}\\ &\rm \approx 352\; L\end{aligned}$.