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

a) 0.38 mol.

b) 0.38 mol.

c) 26.94 g.

Explanation:

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?  

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.