Question

18.74 g of P4 is reacted with 9.6 g of Cl2 balanced is P4 (s) + 6 Cl2 (g) → 4 PCl3 (l) how many PlC3 can I make?

Answer 1

P4(s) + 6Cl2(g) → 4PCl3(l)

n P4 = m P4 / Mr P4

n P4 = 18.74 / 123.88

n P4 = 0.151 mol

n Cl2 = m Cl2 / Mr Cl2

n Cl2 = 9.6 / 70.9

n Cl2 = 0.135 mol

Cl2 act as limiting reactant

n PCl3 = (coef. PCl3)/(coef. Cl2) • n Cl2

n PCl3 = (4/6) • 0.135

n PCl3 = 0.09 mol

m PCl3 = n PCl3 • Mr PCl3

m PCl3 = 0.09 • 137.33

m PCl3 = 12.36 gr