Balance the chemical reaction equation
P4(s)+Cl2(g)→PCl5(g)

Enter the coefficients in order, separated by commas (e.g., 1,2,3).

View Available Hint(s)

1,10,4

The balanced equation is
P4(s)+10Cl2(g)→4PCl5(g)

Calculations involving a limiting reactant

Now consider a situation in which 28.0 g of P4 is added to 53.0 g of Cl2, and a chemical reaction occurs. To identify the limiting reactant, you will need to perform two separate calculations:
Calculate the number of moles of PCl5 that can be produced from 28.0 g of P4 (and excess Cl2).
Calculate the number of moles of PCl5 that can be produced from 53.0 g of Cl2 (and excess P4).
Then, compare the two values. The reactant that produces the smaller amount of product is the limiting reactant.
Part B

How many moles of PCl5 can be produced from 28.0 g of P4 (and excess Cl2)?

Express your answer numerically in moles.

Part C

How many moles of PCl5 can be produced from 53.0 g of Cl2 (and excess P4)?

From the reaction we see that 1 mole of P₄ will react with 10 moles of Cl₂ so 0.226 moles of P₄ will react with 2.226 moles of Cl₂, but we have only 0.746 moles of Cl₂. So the limiting reactant will be Cl₂.

Part B

From the previous calculations we know that 28 g of P₄ are equal to 0.226 moles moles of P₄.

Now, knowing the chemical reaction, we devise the following reasoning:

if 1 mole of P₄ produces 4 moles of PCl₅

then 0.226 moles of P₄ produces X moles of PCl₅

X = (0.226 × 4) / 1 = 0.904 moles of PCl₅

Part C

From the previous calculations we know that 53 g of Cl₂ are equal to 0.746 moles moles of Cl₂.

Now, knowing the chemical reaction, we devise the following reasoning:

if 10 moles of Cl₂ produces 4 moles of PCl₅

then 0.746 moles of Cl₂ produces X moles of PCl₅

X = (0.746 × 4) / 10 = 0.298 moles of PCl₅