Question

Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following balanced equation:

2KI(aq) + Cl2(g) → 2KCl(aq) + I2(g)

How many moles of I2 are produced?
mol
How many moles of Cl2 are used?
mol
How many grams of Cl2 (g) are used?
grams

Answer 1

Assume that 8.5 L of iodine gas (I2) are produced at STP according to the following balanced equation:

2KI(aq) + Cl2(g) → 2KCl(aq) + I2(g)

How many moles of I2 are produced?-----0.38 mol

How many moles of Cl2 are used?----------0.38 mol

How many grams of Cl2 (g) are used? ----

27 grams

Step-by-step explanation:

edge2023

Answer 2

a) 0.38 mol.

b) 0.38 mol.

c) 26.94 g.

Step-by-step explanation:

• For the balanced equation:

2KI(aq) + Cl₂(g) → 2KCl(aq) + I₂(g),

It is clear that 2 mol of KI react with 1 mol of Cl₂ to produce 2 mol of KCl and 1 mol of I₂.

a) How many moles of I₂ are produced?

• Firstly, we need to calculate the no. of moles of 8.5 L of produced I₂:

It is known that every 1.0 mol of any gas occupies 22.4 L at STP conditions.

Using cross multiplication:

1 mol of I₂ occupies → 22.4 L, at STP.

??? mol of I₂ occupies → 8.5 L, at STP.

∴ The no. of moles of I₂ produced = (1 mol)(8.5 mol)/(22.4 L) = 0.38 mol.

b) How many moles of Cl₂ are used?

Using cross multiplication:

1 mol of Cl₂ produces → 1 mol of I₂, from stichiometry.

∴ 0.38 mol of Cl₂ produces → 0.38 mol of I₂.

So, the no. of moles of Cl₂ are used = 0.38 mol.

c) How many grams of Cl₂(g) are used?

∴ The "no. of grams" of Cl₂(g) are used = (no. of moles of Cl₂)(molar mass of Cl₂) = (0.38 mol)(70.9 g/mol) = 26.94 g.