Question

What is the excess reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3Ca(NO3)2 + 2Li3PO4 → 6LiNO3 + Ca3(PO4)2 a) Li3PO4 b) Ca3(PO4)2 c) Ca(NO3)2 d) LiNO3 e) not enough information.

Answer 1

Answer: The excess reactant for the given reaction is $Li_3PO_4$

Explanation:

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

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

For the given chemical reaction:

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow 6LiNO_3+Ca_3(PO_4)_2$

We are given:

Moles of calcium nitrate = 3.4 mol

Moles of lithium phosphate = 2.4 mol

By stoichiometry of the reaction:

3 moles of calcium nitrate reacts with 2 moles of lithium phosphate

So, 3.4 moles of calcium nitrate will react with = $\frac{2}{3}\times 3.4=2.27mol$ of lithium phosphate

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

Thus, calcium nitrate is considered as the limiting reagent because it limits the formation of products.

Hence, the excess reactant for the given reaction is $Li_3PO_4$