6c. Calculate the maximum volume, in dm3, of chlorine gas at Stp that can be obtained from 23.4 tonnes of molten sodium chloride. 1 tonne= 10^6 Mr of NaCl=58.5 molar volume of chlorine at stp= 24dm3 Give your answer in standard format.

4.48×10⁶ dm³

Explanation:

We'll begin by converting 23.4 tonnes to grams (g). This can be obtained as follow:

1 tonne = 10⁶ g

Therefore,

23.4 tonnes = 23.4 × 10⁶

23.4 tonnes = 2.34×10⁷ g

Thus, 23.4 tonnes is equivalent to 2.34×10⁷ g

Next, we shall determine the number of mole in 2.34×10⁷ g of NaCl. This can be obtained as follow:

Mass NaCl = 2.34×10⁷ g

Molar mass of NaCl = 58.5 g/mol

Mole of NaCl =?

Mole = mass / molar mass

Mole of NaCl = 2.34×10⁷ / 58.5

Mole of NaCl = 4×10⁵ moles

Next, we shall determine the number of mole of chlorine, Cl₂ produced from the reaction. This can be obtained as follow:

2NaCl —> 2Na + Cl₂

From the balanced equation above,

2 moles of NaCl reacted to produce 1 mole of Cl₂.

Therefore, 4×10⁵ moles of NaCl will react to produce = (4×10⁵ × 1)/2 = 2×10⁵ moles of Cl₂.

Thus, 2×10⁵ moles of Cl₂ were obtained from the reaction.

Finally, we shall determine the volume of Cl₂ produced. This can be obtained as follow:

1 mole of Cl₂ at stp = 22.4 dm³

Therefore,

2×10⁵ moles of Cl₂ at stp = 2×10⁵ 22.4

2×10⁵ moles of Cl₂ at stp = 4.48×10⁶ dm³

Thus, the volume of chlorine obtained from the reaction is 4.48×10⁶ dm³