Дано m(Ca)=44 g W(прим)=30% ------------------- V(CL2)-? m(CaCL2)-? m(чист Ca)=44-(44*30%/100%)=30.8 g 30.8 y X Ca + CL2-->CaCl2 Vm=22.4L/mol 40 22.4 111 M(Ca)=40 g/mol M(CaCL2)=111 g/mol X(CaCL2)=30.8*111/40 = 85.47 g Y(CL2)=30.8*22.4 / 40= 17.25 L ответ 17.25 л, 85.47 г