What mass of pcl5 will be produced from the given masses of both reactants?

Part A Balance the chemical reaction equation P4(s) +Cl2(g)PCls (g) Enter the coefficients in order, separated by commas (e.g., 1,2,3) View Available Hint(s) 1,10,4 Previous Answers Correct The balanced equation is P4 (s) +10Cl2(g)-4PCI5 (g)
Part B How many moles of PCls can be produced from 29.0 g of P4 (and excess Cl2)? Express your answer to three significant figures and include the appropriate units View Available Hint(s) 0.936 mole Previous Answers Correct Part C How many moles of PCls can be produced from 57.0 g of Cl2 (and excess P4)? Express your answer to three significant figures and include the appropriate units View Available Hint(s) 0.322 mole Previous Answers Correct
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 View Available Hint(s) με Value Units SubmitPr Previous Answers X Incorrect; Try Again; 5 attempts remaining

Answer

Moles of P4 = mass/molar mass

= 29.0/123.895 = 0.234 mole

Moles of Cl2 = 57.0/71 = 0.802817 mole

Now, balanced equation is–

P4(s) + 10Cl2(g) –> 4PCl5(g)

Moles of P4 required to react completely with Cl2 = (1/10)*moles
of Cl2

= (1/10)*0.802817 = 0.0802817 mole

But P4 present = 0.234 mole ( excess)

Limiting reagent is Cl2 then.

Moles of PCl5 produced = (4/10)*moles of Cl2

= (4/10)*0.802817 = 0.32112 mole

Mass of PCl5 = 0.32112*208.24 = 66.87 gram

= 66.9 gram

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