Question

A side reaction in the manufacture of rayon from wood pulp is \(3 \mathrm{CS}_{2}+6 \mathrm{NaOH} \longrightarrow 2 \mathrm{Na}_{2} \mathrm{CS}_{3}+\mathrm{Na}_{2} \mathrm{CO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\) How many grams of \(\mathrm{Na}_{2} \mathrm{CS}_{3}\) are produced in the reaction of \(92.5 \mathrm{mL}\) of liquid \(\mathrm{CS}_{2}(d=1.26 \mathrm{g} / \mathrm{mL})\) and 2.78 mol NaOH?

Short answer

The mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) produced in the reaction is approximately 146.34 g.

Step-by-step solution

Convert volume to mass of CS2

First, convert the volume of \(\mathrm{CS}_{2}\) to mass by using the given density (d = 1.26 g/mL). The formula to calculate mass from volume and density is: \[ \text{Mass} = \text{Density} * \text{Volume} \]. Convert 92.5 mL of \(\mathrm{CS}_{2}\) to grams which is: \( 1.26 \, \mathrm{g/mL} * 92.5 \, \mathrm{mL} = 116.55 \, \mathrm{g} \) of \(\mathrm{CS}_{2}\).

Convert mass to moles of CS2

To solve this problem, we must deal with moles. The formula to find moles from mass and molar mass is: \[ \text{Moles} = \frac{\text{Mass}}{\text{Molar mass}} \]. The molar mass of \(\mathrm{CS}_{2}\) is 76.141 g/mol. Substituting the given values, we derive that there are \( \frac{116.55 \, \mathrm{g}}{76.141 \, \mathrm{g/mol}} = 1.530 \, \mathrm{mol} \) of \(\mathrm{CS}_{2}\).

Use stoichiometric ratios

According to the balanced chemical reaction, 3 moles of \(\mathrm{CS}_{2}\) produces 2 moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\). Thus, we will calculate the moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) that could be produced given the moles of \(\mathrm{CS}_{2}\) and \(\mathrm{NaOH}\) according to these stoichiometric ratios. Based on \(\mathrm{CS}_{2}\), \(\frac{1.530 \, \mathrm{mol} \, \mathrm{CS}_{2}}{3} * 2 = 1.020 \, \mathrm{mol} \, \mathrm{Na}_{2}\mathrm{CS}_{3}\) can be theoretically produced. According to \(\mathrm{NaOH}\), \(\frac{2.78 \, \mathrm{mol} \, \mathrm{NaOH}}{6} * 2 = 0.926 \, \mathrm{mol} \, \mathrm{Na}_{2}\mathrm{CS}_{3}\) can be theoretically produced.

Identify the limiting reactant and calculate the mass of sodium thiocarbonate

Since fewer moles of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) can be formed from \(\mathrm{NaOH}\), NaOH is the limiting reactant. We calculate the mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) that can be formed from this amount of \(\mathrm{NaOH}\) using this formula: \[ \text{Mass} = \text{Moles} * \text{Molar mass} \]. The molar mass of \(\mathrm{Na}_{2}\mathrm{CS}_{3}\) is 158.11 g/mol. Substituting the given values, we derive that \( 0.926 \, \mathrm{mol} * 158.11 \, \mathrm{g/mol} = 146.34 \, \mathrm{g} \, \mathrm{Na}_{2}\mathrm{CS}_{3} \).

Perform check

To make sure everything has been calculated correctly, check the relation between the moles of \(\mathrm{NaOH}\) and the moles of \(\mathrm{CS}_{2}\). As seen from the chemical equation, the mole ratio \(\mathrm{NaOH} : \mathrm{CS}_{2}\) should be 6:3 or 2:1. The given moles of \(\mathrm{NaOH}\) is 2.78 mol, and the moles of \(\mathrm{CS}_{2}\) is 1.530 mol. These values do indeed respect the 2:1 ratio, so our answer is most likely correct.

Key concepts

Limiting reactant

In any chemical reaction, the limiting reactant is the substance that gets completely used up first. This reactant limits the amount of product that can be formed. In our exercise, we start with 92.5 mL of carbon disulfide ( CS_{2} ) and 2.78 mol of sodium hydroxide (NaOH).

To find which reactant is limiting, we must calculate how much product can be formed from each reactant based on the balanced equation. By compiling these calculations, it turns out that NaOH is the limiting reactant. This is because the amount of sodium thiocarbonate ( Na_{2}CS_{3} ) it forms is less compared to using all of CS_{2} .

Knowing the limiting reactant is crucial in stoichiometry because it tells us exactly how much product we can produce. It also helps us understand which reactant is in excess, indicating that not all reactants will be completely used.

Mole ratios

Mole ratios are derived from the coefficients of a balanced chemical equation and tell us how much of one substance reacts with another. In the chemical equation for our problem, 3 moles of CS_{2} reacts with 6 moles of NaOH to give products, including 2 moles of Na_{2}CS_{3} .

This ratio is vital as it allows us to convert moles of one substance to moles of another using simple multiplication or division. For example:

• From the exercise, with 1.530 mol of CS_{2} , we can produce 1.020 mol of Na_{2}CS_{3} by using the ratio directly from the reaction.
• Similarly, with 2.78 mol of NaOH, only 0.926 mol of Na_{2}CS_{3} can be formed, showing up the discrepancies in theoretical production based on available reactants.

This direct application of mole ratios ensures precise predictions of reactants and products in a reaction.

Chemical reactions

Chemical reactions involve the rearrangement of atoms to transform reactants into products. The balanced equation given in the exercise details the reaction between CS_{2} and NaOH producing Na_{2}CS_{3} , Na_{2}CO_{3} , and water.

A reliable balanced equation is key to solving stoichiometric problems:

• It ensures mass conservation, showing equal amounts of atoms for each element in both reactants and products.
• It provides exact ratios needed for mole calculations.

Understanding chemical reactions involves identifying the type of reaction taking place and using the provided details to carry out calculations like determining the limiting reactant or expected product quantities. This comprehensive understanding aids in practical applications, such as predicting outcomes of reactions in industrial settings.