Identify the limiting reactant when 65.14g of CaCl, reacts with 74.68 g of N to produce CaCO, and NaCl.

This is an incomplete question, here is a complete question.

Identify the limiting reactant when 65.14 g of CaCl₂, reacts with 74.68 g of Na₂CO₃ to produce CaCO₃, and NaCl.

The limiting reactant is,

Explanation : Given,

Mass of = 65.14 g

Mass of = 74.68 g

Molar mass of = 111 g/mol

Molar mass of = 106 g/mol

First we have to calculate the moles of and .

and,

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 1 mole of react with 1 mole of

So, 0.587 mole of react with 0.587 mole of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Hence, the limiting reactant is,

the answer is the

i just learnt this recently but i need to look through my notes again