дано
m(al) = 21.6 g
W(CuCL2) = 20%
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m(ppa CuCL2)-?
2Al+3CuCL2-->3Cu+2AlCL3
M(Al) = 27 g/mol
n(Al) = m/M = 21.6 / 27 = 0.8 mol
2n(Al) = 3n(CuCL2)
n(CuCL2) = 3*0.8 / 2 = 1.2 mol
M(CuCL2) = 135 g/mol
m(CuCL2) = n*M = 1.2 * 135 = 162 g
m(ppa CuCL2) = 162*100% / 20% = 810 g
ответ 810 г