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# Answer to Question #55726 in General Chemistry for sue

Question #55726
Part B How many moles of PCl5 can be produced from 26.0 g of P4 (and excess Cl2)? Express your answer to three significant figures and include the appropriate units. 0.839 mole....answer to B Part C How many moles of PCl5 can be produced from 59.0 g of Cl2 (and excess P4)? Express your answer to three significant figures and include the appropriate units. 0.333 mole....answer to C Part D What mass of PCl5 will be produced from the given masses of both reactants? Express your answer to three significant figures and include the appropriate units. **Need part D** Shoud have 3 SF and unit
1
2015-10-22T09:32:33-0400
Solution
P4 + 10 Cl2 = 4 PCl5
Combination of one mole of P4 with ten moles of chlorine gives four moles of PCl5.
MW(P4) = 123.895048 g mol-1
MW(Cl2) = 70.906 g mol-1
MW(PCl5) = 208.22 g mol-1
Part B
n(P4) = m(P4)/MW(P4) = 26.0 g/123.895048 g mol-1 = 0.210 mol
n(PCl5) = 4 n(P4) = 0.839 mol
Part C
N(Cl2) = m(Cl2)/MW(Cl2) = 59.0 g/70.906 g mol-1 = 0.832 mol
n(Cl5) = 0.4 n(Cl2) = 0.333 mol
Part D
Chlorine is limiting reagent. 0.333 moles of PCl5 will be formed:
m(PCl5) = n(PCl5)*MW(PCl5) = 0.333 mol * 208.22 g mol-1 = 69.3 g

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