In: Chemistry
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)?
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).