Question

what is the smallest amount of oxygen molecules needed for the complete combustion of 40 g of methanol?
A. 1.88 moles
B. 2.50 moles
C. 3.75 mole
D. 5.00 moles​

Answer 1

Answer:

A

Explanation:

Answer 2

Answer:

A. 1.88 moles

Explanation:

CH3OH + 3/2 O2  -> CO2 + 2H2O

mr of CH3OH= 32

Mole= Given mass/ molecular mass

Mole of CH3OH= 40/32 = 1.25 mol

Ratio table:  CH3OH          O2

                           1          :        3/2

                         1.25      :          x

x=1.25 x 3/2 = 1.875 moles = 1.88 moles