При нагревании метанола массой 2,4 г с уксусной кислотой получили метилацетат

дано

m(CH3COOH) = 2.4 g

m пр (CH3COOCH3) = 3.7g

----------------------------------

(CH3COOCH3)-?

CH3COOH+CH3OH--gt;CH3COOCH3+H2O

M(CH3COOH) = 60 g/mol

n(CH3COOH)  = m/M = 2.4 / 60 = 0.04 mol

n(CH3COOH) = n(CH3COOCH3)= 0.04 mol

M(CH3COOCH3) = 74 g/mol

mтеор (CH3COOCH3) = n*M = 0.04 * 74 = 2.96 g

(CH3COOCH3) = mеор(CH3COOCH3) / m пр (CH3COOCH3) * 100%

n(CH3COOCH3)  = 2.96 / 3.7 * 100% = 80%

ответ 80%