Answer to Question #311405 in General Chemistry for Victoria

Question

ii) Calculate the concentration of methanol produced by the reaction in b), with the following reactant concentrations:

CO: 2.13 x 10^-3 mol L ^-1. H: 1.4 x 10^-3 mol L^-1

Answer 1

The reaction of the methanol production is following:

$CO+2H_2= CH_3OH$

To obtain the concentration of methanol the limited reagent has to be deduced. For this, the mole numbers of each must be obtained.

$c= \frac{n}{V}$

where c is a concentration, n is a mole number and V is volume. Assuming that the volume of two components is equal to 1L, the concentration becomes equal to the mole number. Since the mole number of hydrogen is less as compared to CO gas, this reagent will react without excess in the methanol formation and must be used for determination of the methanol concentration.

By the reaction, there are two hydrogen molecules participate in the methanol formation, so the mole number i.e. concentration of methanol will be equal to the concentration of hydrogen divided by two:

$c(CH_3OH)=\frac{c(H_2)}{2}=\frac{1.4 \times 10^{-3}}{2}=0.7 \times 10^{-3} mol/L$