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When 3.9 g of Al(OH)3 is reacted with excess of HCl, 6.5 g of AlCl3 is formed. Find the % yield of product.

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When 3.9 g of Al(OH)3 is reacted with excess of HCl, 6.5 g of AlCl3 is formed. Find the % yield of product.

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Al(OH)3 + 3HCl → AlCl3 + 3H2O

78 g Al(OH)3 produce 133.5 g AlCl3

But amount of AlCl3 formed = 6.5 g

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