Дано
m(Ca)=44 g
W(прим)=30%
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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