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
Expert's answer
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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