Answer to Question #180816 in General Chemistry for Sadie Dancy

General Chemistry

Question #180816

How many grams of Al is needed to make 0.59g of AlCl3?

Expert's answer

2Al(s)+3Cl2(g)→2AlCl3(s)

Moles of AlCl3"=\\frac{0.59}{133}=4.4\u00d710^{-3}"

Moles of Al "=4.4\u00d710^{-3}"

grams of Al = "4.4\u00d710^{-3}\u00d727 =0.12g"

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