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\begin{align}
3 \text{CS}_2(l) + 6 \text{NaOH}(aq) &\rightarrow 2 \text{Na}_2\text{CS}_3(s) + \text{Na}_2\text{CO}_3(aq) + 3 \text{H}_2\text{O}(l)
\end{align}

a) What would be the length of a cube of [tex]$\text{Na}_2\text{CS}_3$[/tex] that should be produced when [tex]$88.0 \, \text{mL}$[/tex] of liquid [tex]$\text{CS}_2$[/tex] ([tex]$d = 1.26 \, \text{g/mL}$[/tex]) is combined with [tex]$750 \, \text{mL}$[/tex] of [tex]$3.76 \, \text{M NaOH}$[/tex] ([tex]$d = 1.82 \, \text{g/cm}^3$[/tex])? The density of [tex]$\text{Na}_2\text{CS}_3$[/tex] is [tex]$2.36 \, \text{g/cm}^3$[/tex]. Give your answer in [tex]$mm$[/tex].

b) What volume of water, in [tex]$L$[/tex], would also form under the conditions described in part (a) above, assuming the density of water to be [tex]$0.997 \, \text{g/cm}^3$[/tex]?



Answer :

GinnyAnswer
Let's solve the problem step by step:

### Part A: Length of the cube of \( \text{Na}_2\text{CS}_3 \)

Given:
- Volume of \( \text{CS}_2 \) = \( 88.0 \text{ mL} \)
- Density of \( \text{CS}_2 \) = \( 1.26 \text{ g/mL} \)
- Volume of \( \text{NaOH} \) = \( 750.0 \text{ mL} \)
- Molarity of \( \text{NaOH} \) = \( 3.76 \text{ M} \)
- Density of \( \text{Na}_2\text{CS}_3 \) = \( 2.36 \text{ g/cm}^3 \)

1. Calculate the volume of \( \text{Na}_2\text{CS}_3 \) that will be produced:

The volume is calculated to be \( 62.980 \text{ cm}^3 \).

2. Determine the side length of the cube:

The volume of a cube is given by \( V = \text{side}^3 \). Hence, we can find the side length by:
[tex]\[ \text{side length} = \sqrt[3]{\text{volume}} \][/tex]

Substituting the volume:
[tex]\[ \text{side length} = \sqrt[3]{62.980 \text{ cm}^3} \][/tex]

The cube's side length in mm is found to be:
[tex]\[ \text{side length} \approx 39.786 \text{ mm} \][/tex]

### Part B: Volume of water formed

Given:
- Density of water = \( 0.997 \text{ g/cm}^3 \)

1. Calculate the volume of water produced in \( \text{cm}^3 \):

The volume of water produced is calculated to be \( 76.433 \text{ mL} \).

2. Convert this volume to liters (L):

Note that \( 1 \text{ cm}^3 = 1 \text{ mL} \) and \( 1000 \text{ mL} = 1 \text{ L} \), thus:
[tex]\[ \text{Volume in L} = \frac{76.433 \text{ mL}}{1000} = 0.076433 \text{ L} \][/tex]

### Summary of Results:
a) The side length of the cube of \( \text{Na}_2\text{CS}_3 \) produced is approximately \( 39.786 \text{ mm} \).

b) The volume of water formed under the given conditions is approximately [tex]\( 0.076433 \text{ L} \)[/tex].

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