Question

A volume of \(200 \mathrm{ml}\) of oxygen is added to \(100 \mathrm{ml}\) of a mixture containing \(\mathrm{CS}_{2}\) vapour and \(\mathrm{CO}\), and the total mixture is burnt. After combustion, the volume of the entire mixture is \(245 \mathrm{ml}\). Calculate the volume of the oxygen that remains (a) \(67.5 \mathrm{ml}\) (b) \(125.0 \mathrm{ml}\) (c) \(200.0 \mathrm{ml}\) (d) \(100.0 \mathrm{ml}\)

Short answer

Question: Calculate the remaining volume of oxygen after the combustion of a mixture containing 100 ml of CS2 and CO vapors mixed with 200 ml of oxygen if the total volume after combustion is 245 ml. Answer: The remaining volume of oxygen after the combustion is 145 ml.

Step-by-step solution

Write the balanced chemical equations for the combustion reactions

The combustion reactions for both \(\mathrm{CS}_{2}\) and \(\mathrm{CO}\) are given below: \(\mathrm{CS}_{2} + 3\mathrm{O}_2 \rightarrow \mathrm{CO}_2 + 2\mathrm{SO}_2\) \(2\mathrm{CO} + \mathrm{O}_2 \rightarrow 2\mathrm{CO}_2\) Note: There are different ways to balance the combustion reactions, but this set of equations is balanced and will help in solving the problem.

Calculate the total initial volume of the mixture

Using the given volumes of oxygen and the mixture containing \(\mathrm{CS}_{2}\) and \(\mathrm{CO}\), we can find the total initial volume of the mixture: Initial volume of mixture \(= 200\mathrm{ml} + 100\mathrm{ml}\) (oxygen + \((\mathrm{CS}_{2}+\mathrm{CO})\)) Initial volume of mixture \(= 300\mathrm{ml}\)

Calculate the decrease in volume due to combustion

After combustion, the total volume of the mixture is \(245\mathrm{ml}\). The decrease in volume due to combustion can be calculated as: Decrease in volume \(= 300\mathrm{ml} - 245\mathrm{ml}\) Decrease in volume \(= 55\mathrm{ml}\)

Use stoichiometry to find the remaining volume of oxygen

According to the stoichiometry of the reactions, 1 mole of \(\mathrm{CS}_{2}\) reacts with 3 moles of \(\mathrm{O}_2\), and 2 moles of \(\mathrm{CO}\) react with 1 mole of \(\mathrm{O}_2\). Therefore the decrease in volume due to combustion is either 3 times the moles of \(\mathrm{CS}_{2}\) burnt or half the moles of \(\mathrm{CO}\) burnt. Let \(x\mathrm{ml}\) be the volume of \(\mathrm{CS}_{2}\) burnt, then \(3x\mathrm{ml}\) of \(\mathrm{O}_2\) is consumed in its combustion. Also, let \(y\mathrm{ml}\) be the volume of \(\mathrm{CO}\) burnt, then \(\frac{1}{2}y\mathrm{ml}\) of \(\mathrm{O}_{2}\) is consumed in its combustion. \(3x\mathrm{ml} + \frac{1}{2}y\mathrm{ml} = 55\mathrm{ml}\) (decrease in volume due to combustion) Since the initial volume of the mixture containing \(\mathrm{CS}_{2}\) and \(\mathrm{CO}\) was \(100\mathrm{ml}\), \(x+y=100\). Now, we have two equations with two unknowns, \(x\) and \(y\). We can solve this system of linear equations to determine the value of \(x\) and \(y\). We can first solve for \(x\) from the second equation: \(x=100-y\) Now, substitute the value of x in the first equation: \(3(100-y) + \frac{1}{2}y = 55\) Solving for \(y\): \(300 - 3y + \frac{1}{2}y = 55 \Rightarrow -\frac{5}{2}y=-245 \Rightarrow y=98\) Now, substituting the value of \(y\) in the equation for \(x\), we get \(x=100-98=2\) Therefore, the volume of \(\mathrm{CS}_{2}\) burnt is \(2\mathrm{ml}\), and the volume of \(\mathrm{CO}\) burnt is \(98\mathrm{ml}\). The remaining volume of oxygen can be calculated as: Remaining volume of \(\mathrm{O}_2\) \( = 200\mathrm{ml} - 3\times2\mathrm{ml} - \frac{1}{2}\times98\mathrm{ml}\) Remaining volume of \(\mathrm{O}_2\) \( = 200\mathrm{ml} - 6\mathrm{ml} - 49\mathrm{ml}\) Remaining volume of \(\mathrm{O}_2\) \( = 145\mathrm{ml}\) The remaining volume of oxygen is not available in the given options (a, b, c, or d). This might be an error in the given problem or options. However, following the presented reasoning, the remaining volume of oxygen should be \(145\mathrm{ml}\).