Question

What mass of PCl5 will be produced from the given masses of both reactants? 28.0 g of P4 and 59.0 g of Cl2

Answer 1

REACTION Pa (S) + 10 Cl2(g) → 4 PCIs(g) Acc to equation 1 mole of phosphorous react with 10 moles of chlorine gas and produce 4 mole of 4 PCIs(g) 1 mole of P410 moles of C12 4 moles of 4 PCIs No of moles Pagiven mass of Pa Molar mass of Pa 28 g/124 = 0.225 moles = No of moles of CI2 given mass of chlorine gas Molar mass of Cl2 59/71-0.831 moles i.e 0.225 mols of P4 should require 2.25 mole of CI2 gas but we have only 0.831 moles of CI2 so given 0.225 mole of P4 is excess reagent and 0.831 moles of Cl2 gas is limiting reactant we have consider limiting reactant Clz 0.831 mole of C12 gas require only 0.0831 (1/10 times of C12 mole) moles of P4 and produce 0.3324 (0.0831 X4) moles of PCis i.e means 0.3324 moles of PCI5 produced mass of PCIS produced moles of PCI5 X molar mass of PCIS - 0.3324 X 208 g 69.14 g