If 0.84 mol of CS2 reacts with oxygen completely according to the equation CS2(ℓ) + 3 O2(g) → CO2(g) + 2 SO2(g) what volume (total) would the products occupy if they were measured at STP?

The products will occupy a volume of 56.448 L when measured at STP

Explanation:

The first thing to check here is the mole ratio of the reactant in question to the product.

From the chemical equation, we can see that 1 mole of CS2 gave a total mole of 3 moles of the products(1 mole of carbon iv oxide and 2 moles of sulphur iv oxide)

Now if 1 mole of reactant can give 3 moles of the products,

then 0.84 mole of reactant will give 3 * 0.84 moles of the product = 2.52 moles of the products

We are told to calculate the total volume.

At STP , 1 mole of a gaseous substance will occupy a volume of 22.4L or 22.4 dm^3 (liters and dm^3 are same)

This means that 2.52 moles will occupy a volume of 22.4 * 2.52 = 56.448L