Question

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?

Answer 1

The total volume of the gas products (CO2 and SO2) produced from the complete reaction of 0.84 mol of CS2 with oxygen at STP is approximately 56.48 liters.

Step-by-step explanation:

When 0.84 mol of CS2 reacts completely with oxygen according to the balanced chemical equation CS2(l) + 3 O2(g) → CO2(g) + 2 SO2(g), we have to calculate the total volume of the gas products at STP (Standard Temperature and Pressure).

At STP, one mole of any ideal gas occupies 22.414 liters. According to the stoichiometry of the reaction, 1 mole of CS2 produces 1 mole of CO2 and 2 moles of SO2. Therefore, for 0.84 moles of CS2, we would have:

• 0.84 moles of CO2 × 22.414 L/mol = 18.82776 L of CO2
• 0.84 moles × 2 moles of SO2/mole of CS2 = 1.68 moles of SO2
• 1.68 moles of SO2 × 22.414 L/mol = 37.65552 L of SO2

To determine the total volume of all gases produced, we add the volumes of CO2 and SO2:

Total Volume = 18.82776 L CO2 + 37.65552 L SO2 = 56.48328 L

The total volume of the gas products that would occupy at STP after the reaction is completed is approximately 56.48 liters.

Answer 2

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

Step-by-step 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