# What volume of methanol (density 0.8) is required to obtain 37 g of methyl ester of acetic acid

**What volume of methanol (density 0.8) is required to obtain 37 g of methyl ester of acetic acid, if its yield is 80% of theoretical.**

M (CH3COOCH3) = 74 g / mol.

37 g – 80%

X – 100%,

X = (37g × 100%): 80% = 46.25g

Let’s find the amount of CH3COOSH3 substance by the formula:

n = m: M.

n = 46.25 g: 74 g / mol = 0.625 mol.

Let’s compose the reaction equation, find the quantitative ratios of substances.

CH3OH + CH3COOH → CH3COOCH3 + H2O.

According to the reaction equation, there is CH3OH for 1 mol of CH3COOCH3. Substances are in quantitative ratios 1: 1.

The amount of substance will be the same.

n (CH3OH) = n (CH3COOCH3) = 0.625 mol.

Let’s find the mass of CH3OH by the formula:

m = n × M,

M (CH3OH) = 32 g / mol.

m = 0.625 mol × 32 g / mol = 20 g.

V = m: p.

V = 20 g: 0.8 = 25 ml.

Answer: 25 ml.