Answer to Question #55726 in General Chemistry for sue

Question

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

Answer 1

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