Verified answer

дано

W(HCL) = 15%

p(HCL) = 1.13 g/mL

m(CaCO3*MgCO3) = 115 g

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V(ppa HCL)-?

CaCO3*MgCO3+4HCL-->CaCL2+MgCL2+2H2O+2CO2

M(CaCO3*MgCO3) = 184 g/mol

n(CaCO3*MgCO3) = m/M =115 / 184 = 0.625 mol

n(CaCO3*MgCO3) = 4n(HCL)

n(HCL) = 4*0.625 = 2.5 mol

M(HCL) = 36.5 g/mol

m(HCL) = n*M =2.5 * 36.5 = 91.25 g

m(ppa HCL) = 91.25 * 100% / 15%  = 608.33 g

V(ppa HCL) = m(ppa HCL) / p(HCL) = 608.33 / 1.13  = 538.35 L

ответ 538.35 л