1. What is the percent yield of a reaction in which 217 g of phosphorus trichloride...

1. What is the percent yield of a reaction in which 217 g of phosphorus trichloride reacts with excess water to form 115 g of HCl and aqueous phosphorous acid (H3PO3)?

Solutions

Expert Solution

Write the balanced chemical equation for the reaction.

PCl₃ + 3 H₂O -------> 3 HCl + H₃PO₃

As per the stoichiometric equation,

1 mole PCl₃ = 3 moles HCl.

Molar mass of PCl₃ = (1*30.9734 + 3*35.453) g/mol = 137.3324 g/mol.

Molar mass of HCl = (1*1.008 + 1*35.453) g/mol = 36.461 g/mol.

Find the number of moles of PCl₃ corresponding to 217 g.

Moles of PCl₃ = (mass of PCl₃)/(molar mass of PCl₃) = (217 g)/(137.3324 g/mol) =1.5801 mole.

Moles of HCl formed = (1.5801 mole PCl₃)*(3 moles HCl/1 mole PCl₃) = 4.7403 mole HCl.

Mass of HCl formed = (moles of HCl formed)*(molar mass of HCl) = (4.7403 mole)*(36.461 g/mol) = 172.8361 g.

The above gives the theoretical yield of HCl; however, the reaction yielded 115 g HCl. Therefore, the percent yield of the reaction is (yield obtained)/(theoretical yield)*100 = (115 g)/(172.8361 g)*100 = 66.5370% ≈ 66.54% (ans).

queen_honey_blossom answered 1 year ago

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