Ответы на вопрос:
дано
m(al) = 21.6 g
w(cucl2) = 20%
m(ppa
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
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