Question

How many grams of sodium sulfide are formed if 1.80 g of hydrogen sulfide is bubbled into a solution containing 2.40 g of sodium hydroxide, assuming that the sodium sulfide is made in 93.0 % yield?

Answer 1

Answer: The mass of sodium sulfide formed will be 2.18 grams.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$      .....(1)

• For hydrogen sulfide:

Given mass of hydrogen sulfide = 1.80 g

Molar mass of hydrogen sulfide = 34.1 g/mol

Putting values in equation 1, we get:

$\text{Moles of hydrogen sulfide}=\frac{1.80g}{34.1g/mol}=0.053mol$

• For NaOH:

Given mass of NaOH = 2.40 g

Molar mass of NaOH = 40 g/mol

Putting values in equation 1, we get:

$\text{Moles of NaOH}=\frac{2.40g}{40g/mol}=0.06mol$

The chemical equation for the reaction of hydrogen sulfide and NaOH follows:

$H_2S+2NaOH\rightarrow Na_2S+2H_2O$

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of hydrogen sulfide

So, 0.06 moles of NaOH will react with = $\frac{1}{2}\times 0.06=0.03mol$ of hydrogen sulfide

As, given amount of hydrogen sulfide is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaOH produces 1 mole of sodium sulfide

So, 0.06 moles of NaOH will produce = $\frac{1}{2}\times 0.06=0.03mol$ of sodium sulfide

Now, calculating the mass of sodium sulfide from equation 1, we get:

Molar mass of sodium sulfide = 78.04 g/mol

Moles of sodium sulfide = 0.03 moles

Putting values in equation 1, we get:

$0.03mol=\frac{\text{Mass of sodium sulfide}}{78.04g/mol}\\\\\text{Mass of sodium sulfide}=(0.03mol\times 78.04g/mol)=2.3412g$

To calculate the experimental yield of sodium sulfide, we use the equation:

$\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Percentage yield of sodium sulfide = 93.0 %

Theoretical yield of sodium sulfide = 2.3412 g

Putting values in above equation, we get:

$93.0=\frac{\text{Experimental yield of sodium sulfide}}{2.3412g}\times 100\\\\\text{Experimental yield of sodium sulfide}=\frac{93.0\times 2.3412}{100}=2.18g$

Hence, the mass of sodium sulfide formed will be 2.18 grams.

Answer 2

The reaction of sodium sulfide with hydrogen sulfide can be shown as

$H_2S + 2 NaOH = Na_2S + 2 H_2O \\ Molar mass of H_2S = 34 g /mol\\ Mole of H_2S = mass /molar mass\\ = 1.8 /34 = 0.051 mole\\ Mole of NaOH = 2 * 0.051 = 0.102 mole\\ Theoritical mole of NaOH = mass /molar mass\\ = 2.4 /40 = 0.06 mole$

Sodium hydroxide is completely reacting and thus act as limiting reactant

$Hence, mole of Na_2S required = 0.06 /2 = 0.03 mole\\ Mass of Na_2S = 0.03 * 78 = 2.31 g$

Thus, 2.31 g Na2S produced.