Answer:
Assuming the product is AlCl3
Assuming its 1008 g of metalic Al
It will yeld 4234 g of aluminium chloride.
Explanation:
Al molar mass is 26.98 g/mol, so in 1008 g of Al there is 37.36 mols of Al
The stoichometry ratio between Al and AlCl3 is 1:1, so 37.36 mols of Al would form 37.36 mols of AlCl3 if the reaction were complete, but only 85% yelds, so 1008 g of Al makes up to 0.85x37.36 mols of AlCl3 (31.76 moles)
AlCl3 molar mas is 133.33 g/mol, so 31.76 moles has 4234 g
