Question

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

Answer 1

Given `n=(PV)/(RT)`, there are approx. `1` `mol`.

Explanation:

`n=(PV)/(RT)=((702*mm*Hg)/(760*mm*Hg*atm^-1)xx25.0*L)/(0.0821*L*atm*K^-1*mol^-1xx295.5*K)` `~=`

`1*mol`

This value makes sense, in that we are close to `1*atm`, and `298*K` in the given conditions.

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

`"What was the gas evolved during the fermentation?"`

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