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

To determine how many grams of PCl3 can be made from 18.74 g of P4 and 9.6 g of Cl2, convert the masses of the reactants to moles, identify the limiting reagent (Cl2), and then calculate the theoretical yield of PCl3, which is 12.36 g in this case.

Step-by-step explanation:

To find out how many moles of PCl3 can be made when 18.74 g of P4 reacts with 9.6 g of Cl2 according to the balanced chemical equation P4 (s) + 6 Cl2 (g) → 4 PCl3 (l), we need to perform stoichiometric calculations.

First, calculate the moles of each reactant by using their respective molar masses:

The molar mass of P4 is 123.895 g/mol, so 18.74 g P4 ÷ 123.895 g/mol = 0.151 mol P4.

The molar mass of Cl2 is 70.90 g/mol, so 9.6 g Cl2 ÷ 70.90 g/mol = 0.135 mol Cl2.

According to the balanced equation, 1 mole of P4 reacts with 6 moles of Cl2 to produce 4 moles of PCl3. Hence, Cl2 is the limiting reagent since it is available in fewer moles compared to the stoichiometric requirement. We can now calculate the theoretical yield of PCl3 based on the moles of Cl2:

6 moles Cl2 produce 4 moles PCl3.

0.135 mol Cl2 will produce (0.135 mol Cl2 × (4 moles PCl3 / 6 moles Cl2)) = 0.090 mol PCl3.

Multiply the moles of PCl3 by its molar mass (137.33 g/mol) to find the mass of PCl3 that can be produced:

0.090 mol PCl3 × 137.33 g/mol = 12.36 g of PCl3.

Answer 2

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