дано
m(ZnCL2) = 100 g
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m(AgCL)-?
ZnCL2+2AgNO3-->2AgCL+Zn(NO3)2
M(ZnCL2) = 136 g/mol
n(ZnCL2) = m/M = 100 / 136 = 0.735 mol
n(ZnCL2) = 2n(AgCL)
n(AgCL) = 2*0.735 = 1.47 mol
M(AgCL) = 143.5 g/mol
m(AgCL) = n*M = 1.47 * 143.5 = 211 g
ответ 211 г