Question

Part A

Balance the chemical reaction equation

P4(s)+10Cl2(g)→4PCl5(g)

Part B

How many moles of PCl5 can be produced from 24.0 g of P4 (and excess Cl2)?

0.775 mole    thats the answer Part C How many moles of PCl5 can be produced from 50.0 g of Cl2 (and excess P4)? Express your answer to three significant figures and include the appropriate units. 0.282 mole thats the answer Part D What mass of PCl5 will be produced from the given masses of both reactants? ​I NEED HELP ON THIS ONE ... PLEASE HELP ME ON D

Answer 1

part-A

P4(s)+10Cl2(g)→4PCl5(g)

part-B

P4(s)+10Cl2(g)→4PCl5(g)

no of moles of P4   = W/G.M.Wt

                               = 24/124   = 0.194 moles

from balanced equation

1 mole of P4 react with Cl2 to gives 4 moles of PCl5

0.194 moles of P4 react with Cl2 to gives = 4*0.194/1   = 0.776 moles of PCl5

part-C

no of moles of Cl2 = w/G.M.Wt

                               = 50/71   = 0.704 moles

P4(s)+10Cl2(g)→4PCl5(g)

10 moles of Cl2 react with P4 to gives 4 moles of PCl5

0.704 moles of Cl2 react with P4 to gives = 4*0.704/10   = 0.2816 moles of PCl5

part-D

from balanced equation

1 mole of P4 react with Cl2 to gives 4 moles of PCl5

0.194 moles of P4 react with Cl2 to gives = 4*0.194/1   = 0.776 moles of PCl5

mass of PCl5 =no of moles * gram molar mass

                      = 0.776*208.5   = 161.786g

10 moles of Cl2 react with P4 to gives 4 moles of PCl5

0.704 moles of Cl2 react with P4 to gives = 4*0.704/10   = 0.282 moles of PCl5

mass of PCl5 =no of moles * gram molar mass

                      = 0.282*208.5   = 58.7136g