Ответ:

дано

W(CaCO3) = 82%

m(CH3COOH) = 25.6 g

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

m(ppa CaCO3)-?

CaCO3+2CH3COOH-->(CH3COO)2Ca+CO2+H2O

M(CH3COOH) =60 g/mol

n(CH3COOH) = m/M = 25.6 / 60 = 0.43 mol

n(CaCO3) = 2n(CH3COOH)

n(CaCO3) = 0.43 / 2 = 0.215 mol

M(CaCO3) = 100 g/mol

m(CaCO3) = n*M = 0.215 * 100 = 21.5 g

m(ppa CaCO3) = m(CaCO3) * 100% / W(CaCO3)  = 21.5 * 100% / 82% = 26.2 g

ответ 26.2 г

Объяснение: