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10 mL of liquid carbon disulphide (speci...

10 mL of liquid carbon disulphide (specific gravity 2.63) is burnt in oxygen. Find the volume of the resulting gases measured at STP.

A

23.25 L

B

22.26 L

C

23.50 L

D

20.08 L

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem step by step, we will follow these calculations: ### Step 1: Calculate the Density of Carbon Disulfide (CS2) Given the specific gravity of carbon disulfide (CS2) is 2.63, we can find its density using the formula: \[ \text{Density of CS2} = \text{Specific Gravity} \times \text{Density of Water} \] Since the density of water is 1 g/mL: \[ \text{Density of CS2} = 2.63 \times 1 \text{ g/mL} = 2.63 \text{ g/mL} \] ### Step 2: Calculate the Mass of Carbon Disulfide Using the density calculated above, we can find the mass of 10 mL of CS2: \[ \text{Mass of CS2} = \text{Density} \times \text{Volume} = 2.63 \text{ g/mL} \times 10 \text{ mL} = 26.3 \text{ g} \] ### Step 3: Determine the Molar Mass of Carbon Disulfide The molar mass of CS2 is given as: \[ \text{Molar Mass of CS2} = 76 \text{ g/mol} \] ### Step 4: Calculate the Volume of Gases Produced at STP From the combustion reaction of CS2 with oxygen, we can write the balanced equation: \[ \text{CS2} + 3\text{O2} \rightarrow \text{CO2} + 2\text{SO2} \] From the reaction, 1 mole of CS2 produces 1 mole of CO2 and 2 moles of SO2, totaling 3 moles of gas. At STP, 1 mole of gas occupies 22.4 L. Therefore, the total volume of gases produced from 1 mole of CS2 is: \[ \text{Volume of gases} = 3 \times 22.4 \text{ L} = 67.2 \text{ L} \] ### Step 5: Calculate the Volume of Gases Produced from 26.3 g of CS2 Using the mass of CS2 we calculated: \[ \text{Volume of gases from 76 g of CS2} = 67.2 \text{ L} \] To find the volume produced from 26.3 g of CS2: \[ \text{Volume of gases from 1 g of CS2} = \frac{67.2 \text{ L}}{76 \text{ g}} \approx 0.8842 \text{ L/g} \] Now, for 26.3 g of CS2: \[ \text{Volume of gases} = 0.8842 \text{ L/g} \times 26.3 \text{ g} \approx 23.25 \text{ L} \] ### Final Answer The volume of the resulting gases measured at STP is approximately: \[ \text{Volume} = 23.25 \text{ L} \]
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