Question

P4(s) + 6Cl2(g) ==> 4PCl3(l)

If 112g of P4 and 303g of Cl2 react what mass of PCl3 is expected to form? What is the limiting reactant?

Answer 1

Number of moles of P4 = 112 g / 123.90 g/mol = 0.904 mole

Number of moles of Cl2 = 303 g / 70.9060 g/mol = 4.27 mole

From the balanced equation we can say that

1 mole of P4 requires 6 mole of Cl2 so

0.904 mole of P4 will require

= 0.904 mole of P4 *(6 mole of Cl2 / 1 mole of P4)

= 5.42 mole of Cl2

But we have only 4.27 mole of Cl2 which is in short so Cl2 is limiting reactant

From the balanced equation we can say that

6 mole of Cl2 produces 4 mole of PCl3 so

4.27 mole of Cl2 will produce

= 4.27 mole of Cl2 *(4 mole of PCl3 / 6 mole of Cl2)

= 17.1 mole of PCl3

mass of 1 mole of PCl3 = 137.33 g so

the mass of 17.1 mole of PCl3 = 2348 g

Therefore, the mass of PCl3 produced would be 2348 g