Помогите срочно 3,4
В 4 там 8%прилили избыток раствора

3)CaCO3-->CaO+CO2CaO+2HNO3-->Ca(NO3)2+H2OCa(NO3)2+H2SO4-->CaSO4+2HNO3Ca⁺²+2NO⁻²₃+2H⁺+SO⁻²₄-->CaSO4+2H⁺+2NO⁻²₃Ca⁺²+SO⁻²₄-->CaSO44)даноm(ppa Al2(SO4)3) = 68.4 gW(Al2(SO4)2) = 8%-------------------------------m(BaSO4)-?m(Al2(SO4)3 = 68.4 * 8% / 100% = 5.472 gAl2(SO4)3 + 3BaCL2-->3BaSO4+2AlCL3M(Al2(SO4)3) = 342 g/moln(Al2(SO4) 3) = m/M = 5.472 / 342 = 0.016 moln(Al2(SO4)3) = 3n(BaSO4)n(BaSO4) = 3*0.016  = 0.048 molM(BaSO4) = 233 g/molm(BaSO4) = n*M = 0.048 * 233 = 11.184 gответ 11.184 г