Question

How many grams of carbon disulfide are needed to completely consume 25.7 L of chlorine gas according to the following reaction at 25 °C and 1 atm?carbon disulfide (s) + chlorine (g)-carbon tetrachloride (l) + sulfur dichloride ( s )_______grams carbon disulfide

Answer 1

Step-by-step explanation

Given:

Volume of chlorine gas, V = 25.7 L,

Pressure, P = 1 atm, and

Temperature, T = 25 °C (25 + 273.15 = 295.15 K)

What to find:

The grams of carbon disulfide are needed to completely consume 25.7 L of chlorine gas.

Step-by-step solution

The question combines using (a) the ideal gas law, and (b) a stoichiometry calculation.

The first step is to write the balanced chemical equation for reaction as shown below:

CS₂ (s) + 4Cl₂ (g) --> CCl₄ (l) + 2SCl₂ (s)

The second step is using the ideal gas law to calculate the number of moles of Cl₂ gas as follows:

`\begin{gathered} PV=nRT \\ \\ n=PV/RT \\ \\ n=\frac{1atm*25.7\text{ }L}{0.0821\text{ }L\text{ }atm\text{ }mol^(-1)\text{ }K^(-1)*295.15\text{ }K} \\ \\ n=1.06\text{ }mol\text{ }Cl_2 \end{gathered}`

The next step is using the balanced chemical reaction and stoichiometry to convert moles of Cl₂ to moles of CS₂.

From the reaction, you can see that 4 moles of Cl₂ react with 1 mole of CS₂. Therefore, 1.06 moles Cl₂ will react with:

`\frac{1.06\text{ }mol\text{ }Cl₂*1\text{ }mol\text{ }CS₂}{4\text{ }mol\text{ }Cl₂}=0.265\text{ }mol\text{ }CS₂`

The final step is to convert 0.265 mol CS₂ to grams CS₂ using its molar mass (76.139 g/mol).

1 mol CS₂ = 76.139 grams CS₂

Therefore 0.265 mol CS₂ will be equal

`\frac{0.265\text{ }mol\text{ }CS₂*76.139\text{ }grams\text{ }CS₂}{1\text{ }mol\text{ }CS₂}=20.18\text{ }grams\text{ }CS₂`

The answer is: 20.18 grams carbon disulfide