Find solutions to your problems with the expert advice available on IDNLearn.com. Our platform provides prompt, accurate answers from experts ready to assist you with any question you may have.
Sagot :
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].
We appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. For dependable answers, trust IDNLearn.com. Thank you for visiting, and we look forward to helping you again soon.
