For the reaction: Al_2(SO_4), + 3BaC_2 rightarrow 3BaSO_4 + 2AlCl, What is the t

Question

For the reaction: Al_2(SO_4), + 3BaC_2 rightarrow 3BaSO_4 + 2AlCl, What is the theoretical yield of aluminum chloride, if 6.00 mol of barium chloride are reacted with an excess of aluminum sulfate? For the reaction: Fe_2O + 3H rightarrow 2Fe + 3H_2O_omega The reaction is run, and 4.50g of Fe_2O is reduced with an excess of H_2(g). The reaction produces 2.60g of iron metal. Calculate the percent yield. The formula molar mass of Fe_2O_3 is 160g/mol. A fuel sample contains 87.4% nitrogen and 12.6% hydrogen, by weight. The molecular weight is 32.05g. What is the empirical formula for this fuel? What is the molecular formula for the fuel?

Explanation / Answer

(a)

When 3 moles of BaCl2 is reacted 2 moles of AlCl3 is formed

So, when 3 moles of BaCl2 is reacted 4 moles of AlCl3 will be formed.

1 mole AlCl3 weighs = 133.34 g

So 3 moles weighs = 3 x 133.34 g == 400 grams

So theroritical yield is 400 g

(b)

Moles of Fe2O3 reacted = 4.50 g /160 g/mol ==0.028125 mol

1 mole Fe2O3 forms 2 moles Fe .

So moles of Fe should be 2 x 0.028125 mol ==0.05625 mol

Mass of iron formed = 0.05624 mol x 55.845 ==3.1412 grams

Actual mass formd = 2.60 grams

% yield = 2.60 / 3.1412 x 100

= 82.8

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