Question

If 16.29 grams of Na2SO4 is mixed with 3.697 grams of C and allowed to react according to the balanced equation: Na2SO4(aq) + 4 C(s) → Na2S(s) + 4 CO(g) What is the limiting reagent?

Answer 1

Answer: The limiting reagent in the given chemical reaction is carbon metal.

Explanation:

Excess reagent is defined as the reagent which is present in large amount in a chemical reaction.

Limiting reagent is defined as the reagent which is present in small amount in a chemical reaction. Formation of product depends on the limiting reagent.

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

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

• For sodium sulfate:

Given mass of sodium sulfate = 16.29 g

Molar mass of sodium sulfate = 142 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium sulfate}=\frac{16.29g}{142g/mol}=0.115mol$

• For carbon:

Given mass of carbon = 3.697 g

Molar mass of carbon = 12 g/mol

Putting values in equation 1, we get:

$\text{Moles of carbon}=\frac{3.697g}{12g/mol}=0.31mol$

For the given chemical reaction:

$Na_2SO_4(aq.)+4C(s)\rightarrow Na_2S(s)+4CO(g)$

By Stoichiometry of the reaction:

4 moles of carbon reacts with 1 mole of sodium sulfate

So, 0.31 moles of carbon will react with = $\frac{1}{4}\times 0.31=0.0775mol$ of sodium sulfate

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

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

Hence, the limiting reagent in the given chemical reaction is carbon metal.