Question

PLEASE DESPERATE HELP! 65 POINTS

A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess O2 gives 24.0 g of a mixture of CO2 and SO2.
Find the mass of sulfur in the original mixture.
Express your answer to two significant figures and include the appropriate units.

Answer 1

The mass of sulfur in the original mixture, determined from the product formed (SO2) during complete combustion, is approximately 12.03 g.

To find the mass of sulfur in the original mixture, we need to determine the mass of sulfur in the product (SO2) formed during complete combustion. Then, we can subtract this from the total mass of the mixture.

1. Find the moles of SO2 produced:

`\[ \text{Moles of } SO2 = \frac{\text{Mass of } SO2}{\text{Molar mass of } SO2} \]`

The molar mass of SO2 is
`\(32.07 \, \text{g/mol} + 2 * 16.00 \, \text{g/mol} = 64.07 \, \text{g/mol}\)`.

`\[ \text{Moles of } SO2 = \frac{24.0 \, \text{g}}{64.07 \, \text{g/mol}} \approx 0.375 \, \text{mol} \]`

2. Since the ratio of moles of SO2 to moles of S in the original mixture is 1:1 (from the balanced chemical equation), the moles of sulfur are also 0.375.

3. Find the mass of sulfur in the original mixture:

`\[ \text{Mass of sulfur} = \text{Moles of sulfur} * \text{Molar mass of sulfur} \]`

The molar mass of sulfur is
`\(32.07 \, \text{g/mol}\)`.

`\[ \text{Mass of sulfur} = 0.375 \, \text{mol} * 32.07 \, \text{g/mol} \approx 12.03 \, \text{g} \]`

So, the mass of sulfur in the original mixture is approximately
`\(12.03 \, \text{g}\)`.