# A gas evolved during the fermentation of sugar was collected at 22.5°C and 702 mmHg. After purification its volume found to be 25.0 L. How many moles of gas were was collected? R = 0.0821 L*atm/K*mol

Dec 10, 2016

Given $n = \frac{P V}{R T}$, there are approx. $1$ $m o l$.

#### Explanation:

$n = \frac{P V}{R T} = \frac{\frac{702 \cdot m m \cdot H g}{760 \cdot m m \cdot H g \cdot a t {m}^{-} 1} \times 25.0 \cdot L}{0.0821 \cdot L \cdot a t m \cdot {K}^{-} 1 \cdot m o {l}^{-} 1 \times 295.5 \cdot K}$ $\cong$

$1 \cdot m o l$

This value makes sense, in that we are close to $1 \cdot a t m$, and $298 \cdot K$ in the given conditions.

By the way? I forgot to ask something that you should know.

$\text{What was the gas evolved during the fermentation?}$

C_6H_12O_6+O_2 stackrel("yeast")rarr"Gas"+"ethanol"+"Water"