Question

) If 23 g of FeCl2 reacts with 41 grams of Na3PO4, what is the limiting reagent? How much NaCl can be formed?

Answer 1

Answer: The limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 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 iron (II) chloride:

Given mass of iron (II) chloride = 23 g

Molar mass of iron (II) chloride = 126.8 g/mol

Putting values in equation 1, we get:

$\text{Moles of iron (II) chloride}=\frac{23g}{126.8g/mol}=0.181mol$

• For sodium phosphate:

Given mass of sodium phosphate = 41 g

Molar mass of sodium phosphate = 164 g/mol

Putting values in equation 1, we get:

$\text{Moles of sodium phosphate}=\frac{41g}{164g/mol}=0.25mol$

The chemical equation for the reaction of iron (II) chloride and sodium phosphate follows:

$3FeCl_2+2Na_3PO_4\rightarrow 6NaCl+Fe_3(PO_4)_2$

By Stoichiometry of the reaction:

3 moles of iron (II) chloride reacts with 2 moles of sodium phosphate

So, 0.181 moles of iron (II) chloride will react with = $\frac{2}{3}\times 0.181=0.1206mol$ of sodium phosphate

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

Thus, iron (II) chloride is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of iron (II) chloride produces 6 mole of sodium chloride.

So, 0.181 moles of iron (II) chloride will produce = $\frac{6}{3}\times 0.181=0.362moles$ of sodium chloride.

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

Molar mass of sodium chloride = 58.5 g/mol

Moles of sodium chloride = 0.362 moles

Putting values in equation 1, we get:

$0.362mol=\frac{\text{Mass of sodium chloride}}{58.5g/mol}\\\\\text{Mass of sodium chloride}=(0.362mol\times 58.5g/mol)=21.2g$

Hence, the limiting reagent is iron (II) chloride and the mass of sodium chloride formed is 21.2 grams