Question

What volume of Cl2 gas, measured at 688 torr and 36 °C, is required to form 22 g of NaCl?Express your answer using two significant figures.

Answer 1

Volume of the gas 5.3L

EXPLANATION;

Given that

The pressure of the chlorine gas is 688 torr

The temperature of the chlorine gas is 36 degrees Celcius

The mass of NaCl is 22 grams

Follow the steps below to find the volume of chlorine gas

Step 1; Write a balanced equation for the reaction

`\text{ 2Na + Cl}_2\text{ }\rightarrow\text{ 2NaCl}`

In the above reaction, 2 moles of Na react with 1 mole of chlorine to form 2 moles of NaCl

Step 2; Find the number of moles of NaCl using the below formula

`\text{ mole = }\frac{\text{ mass}}{\text{ molar mass}}`

Recall, that the molar mass of NaCl is 58.44 g/mol

`\begin{gathered} \text{ mole = }\frac{22}{\text{ 58.44}} \\ \text{ moles = 0.376 mole} \end{gathered}`

Step 3; Find the number of moles of Cl2 using a stoichiometry ratio

Let x represents the number of moles of Cl2

`\begin{gathered} 1\text{ mole Cl}_2\text{ }\rightarrow\text{ 2 moles NaCl} \\ \text{ x moles Cl}_2\rightarrow\text{ 0.376 moles NaCl} \\ \text{ cross multiply} \\ \text{ x moles Cl}_2\text{ }*2\text{ moles NaCl }=\text{ 1 mole Cl}_2\text{ }*\text{ 0.376 mole NaCl} \\ \text{ Isolate x} \\ \text{ x = }\frac{\text{ 1 mole Cl}_2*0.376moles\cancel{NaCl}}{2moles\cancel{NaCl}} \\ \\ \text{ x = }\frac{1\text{ }*\text{ 0.376}}{2} \\ x\text{ = }(0.376)/(2) \\ \text{ x = 0.188 mol} \end{gathered}`

The number of moles of Cl2 is 0.188 mol

Step 4; Find the volume of the gas using the ideal gas equation

`\text{ PV = nRT}`

Convert the temperature to degree kelvin

`\begin{gathered} \text{ t = 36}\degree C \\ \text{ T = 36 + 273.15} \\ \text{ T = 309.15 K} \end{gathered}`

Step 5; Find the volume of the gas

`\begin{gathered} \text{ PV = nRT} \\ \text{ Recall, that R = 62.363 L. torr mol}^(-1)K^(-1) \\ \text{ 688}*\text{ V =}^{\text{ }}0.188\text{ x 62.636 }*\text{ 309.15} \\ \text{ 688 V = 3624. 500} \\ V\text{ = }\frac{\text{ 3645, 500}}{688} \\ \text{ v = 5.3 L} \end{gathered}`

Therefore, volume of the gas 5.3L