22.0 grams of solid iron (II) sulfide reacts with 185 grams of hydrochloric acid (HCl). Determine what products are formed....?

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2015-04-17 1:59 am

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22.0 grams of solid iron (II) sulfide reacts with 185 grams of hydrochloric acid (HCl). Determine what products are formed, balance the chemical equation, determine the limiting reactant and calculate the theoretical yield, in grams, of the product containing sulfur.

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Roger the Mole

2015-04-18 4:53 am

FeS + 2 HCl → FeCl2 + H2S

(22.0 g FeS) / (87.910 g FeS/mol) = 0.25026 mol FeS
(185 g HCl) / (36.4611 g HCl/mol) = 5.0739 mol HCl

It shouldn't take any more arithmetic to see that HCl is in large excess, leaving FeS to be the limiting reactant.

(0.25026 mol FeS) x (1 mol H2S / 1 mol FeS) x (34.0809 g H2S/mol) = 8.53 g H2S in theory

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