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

1 Answer
Dec 10, 2016

Answer:

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"#