Answer :
Sure, let's break this down step by step to solve the given problem:
1. Calculate the number of moles of \(CS_2\):
We have the volume of \(CS_2\) which is 88.0 mL and the density of \(CS_2\) which is 1.26 g/mL.
First, find the mass of \(CS_2\):
[tex]\[ \text{mass of } CS_2 = \text{density} \times \text{volume} = 1.26 \, \text{g/mL} \times 88.0 \, \text{mL} = 110.88 \, \text{g} \][/tex]
Next, calculate the number of moles of \(CS_2\) using its molar mass \( (76.14 \, \text{g/mol}) \):
[tex]\[ \text{moles of } CS_2 = \frac{\text{mass of } CS_2}{\text{molar mass of } CS_2} = \frac{110.88 \, \text{g}}{76.14 \, \text{g/mol}} \approx 1.456 \, \text{mol} \][/tex]
2. Calculate the number of moles of \(NaOH\):
We have the volume of \(NaOH\) which is 750.0 mL and the molarity of \(NaOH\) which is 3.76 M.
Convert the volume of \(NaOH\) to liters:
[tex]\[ \text{volume of } NaOH \text{ in liters} = \frac{750.0 \, \text{mL}}{1000.0} = 0.750 \, \text{L} \][/tex]
Calculate the number of moles of \(NaOH\):
[tex]\[ \text{moles of } NaOH = \text{molarity} \times \text{volume in liters} = 3.76 \, \text{M} \times 0.750 \, \text{L} \approx 2.82 \, \text{mol} \][/tex]
3. Use Stoichiometry to find moles of \(Na_2CS_3\):
According to the balanced equation:
[tex]\[ 3 \, \text{moles of } CS_2 \text{ reacts with } 6 \, \text{moles of } NaOH \text{ to produce } 2 \, \text{moles of } Na_2CS_3 \][/tex]
The moles of \(Na_2CS_3\) can be found using the ratio:
[tex]\[ \text{moles of } Na_2CS_3 = \frac{2}{3} \times \text{moles of } CS_2 = \frac{2}{3} \times 1.456 \, \text{mol} \approx 0.971 \, \text{mol} \][/tex]
4. Calculate the mass of \(Na_2CS_3\) produced:
Use the molar mass of \(Na_2CS_3\) which is 158.1 g/mol:
[tex]\[ \text{mass of } Na_2CS_3 = \text{moles of } Na_2CS_3 \times \text{molar mass of } Na_2CS_3 = 0.971 \, \text{mol} \times 158.1 \, \text{g/mol} \approx 153.49 \, \text{g} \][/tex]
5. Calculate the volume of \(Na_2CS_3\) produced:
Using the density of \(Na_2CS_3\) which is 2.36 g/cm³:
[tex]\[ \text{volume of } Na_2CS_3 = \frac{\text{mass of } Na_2CS_3}{\text{density of } Na_2CS_3} = \frac{153.49 \, \text{g}}{2.36 \, \text{g/cm}^3} \approx 65.04 \, \text{cm}^3 \][/tex]
6. Calculate the side length of the cube:
For a cube, the volume \(V\) is given by the side length \(a\) cubed:
[tex]\[ V = a^3 \][/tex]
Therefore, the side length \(a\) in cm is:
[tex]\[ a = \sqrt[3]{\text{volume}} = \sqrt[3]{65.04 \, \text{cm}^3} \approx 4.022 \, \text{cm} \][/tex]
7. Convert the side length from cm to mm:
Since 1 cm = 10 mm:
[tex]\[ \text{side length in mm} = 4.022 \, \text{cm} \times 10 = 40.22 \, \text{mm} \][/tex]
Thus, the length of a cube of [tex]\(Na_2CS_3\)[/tex] that should be produced is approximately [tex]\(40.22 \, \text{mm}\)[/tex].
1. Calculate the number of moles of \(CS_2\):
We have the volume of \(CS_2\) which is 88.0 mL and the density of \(CS_2\) which is 1.26 g/mL.
First, find the mass of \(CS_2\):
[tex]\[ \text{mass of } CS_2 = \text{density} \times \text{volume} = 1.26 \, \text{g/mL} \times 88.0 \, \text{mL} = 110.88 \, \text{g} \][/tex]
Next, calculate the number of moles of \(CS_2\) using its molar mass \( (76.14 \, \text{g/mol}) \):
[tex]\[ \text{moles of } CS_2 = \frac{\text{mass of } CS_2}{\text{molar mass of } CS_2} = \frac{110.88 \, \text{g}}{76.14 \, \text{g/mol}} \approx 1.456 \, \text{mol} \][/tex]
2. Calculate the number of moles of \(NaOH\):
We have the volume of \(NaOH\) which is 750.0 mL and the molarity of \(NaOH\) which is 3.76 M.
Convert the volume of \(NaOH\) to liters:
[tex]\[ \text{volume of } NaOH \text{ in liters} = \frac{750.0 \, \text{mL}}{1000.0} = 0.750 \, \text{L} \][/tex]
Calculate the number of moles of \(NaOH\):
[tex]\[ \text{moles of } NaOH = \text{molarity} \times \text{volume in liters} = 3.76 \, \text{M} \times 0.750 \, \text{L} \approx 2.82 \, \text{mol} \][/tex]
3. Use Stoichiometry to find moles of \(Na_2CS_3\):
According to the balanced equation:
[tex]\[ 3 \, \text{moles of } CS_2 \text{ reacts with } 6 \, \text{moles of } NaOH \text{ to produce } 2 \, \text{moles of } Na_2CS_3 \][/tex]
The moles of \(Na_2CS_3\) can be found using the ratio:
[tex]\[ \text{moles of } Na_2CS_3 = \frac{2}{3} \times \text{moles of } CS_2 = \frac{2}{3} \times 1.456 \, \text{mol} \approx 0.971 \, \text{mol} \][/tex]
4. Calculate the mass of \(Na_2CS_3\) produced:
Use the molar mass of \(Na_2CS_3\) which is 158.1 g/mol:
[tex]\[ \text{mass of } Na_2CS_3 = \text{moles of } Na_2CS_3 \times \text{molar mass of } Na_2CS_3 = 0.971 \, \text{mol} \times 158.1 \, \text{g/mol} \approx 153.49 \, \text{g} \][/tex]
5. Calculate the volume of \(Na_2CS_3\) produced:
Using the density of \(Na_2CS_3\) which is 2.36 g/cm³:
[tex]\[ \text{volume of } Na_2CS_3 = \frac{\text{mass of } Na_2CS_3}{\text{density of } Na_2CS_3} = \frac{153.49 \, \text{g}}{2.36 \, \text{g/cm}^3} \approx 65.04 \, \text{cm}^3 \][/tex]
6. Calculate the side length of the cube:
For a cube, the volume \(V\) is given by the side length \(a\) cubed:
[tex]\[ V = a^3 \][/tex]
Therefore, the side length \(a\) in cm is:
[tex]\[ a = \sqrt[3]{\text{volume}} = \sqrt[3]{65.04 \, \text{cm}^3} \approx 4.022 \, \text{cm} \][/tex]
7. Convert the side length from cm to mm:
Since 1 cm = 10 mm:
[tex]\[ \text{side length in mm} = 4.022 \, \text{cm} \times 10 = 40.22 \, \text{mm} \][/tex]
Thus, the length of a cube of [tex]\(Na_2CS_3\)[/tex] that should be produced is approximately [tex]\(40.22 \, \text{mm}\)[/tex].
