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дано

m(ppa CuCL2) = 135 g

W(CuCL2) = 8%

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m(AL)-?

m(CuCL2) = m(ppa CuCL2) * W(CuCL2) / 100% = 135*8% / 100%= 10.8 g

2Al+3CuCL2 = 2AlCL3+3Cu

M(CuCL2) = 135 g/mol

n(CuCL2) = m/M = 10.8 / 135 = 0.08 mol

2n(AL) = 3n(CuCL2)

n(AL)= 3* 0.08 / 2 = 0.12 mol

M(Al) = 27 g/mol

m(Al) = n(Al)*M(AL) = 3.24 g

ответ 3.24 г