Question

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

Answer 1

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.

Answer: The limiting reactant is,
`CaCl_2`

Explanation : Given,

Mass of
`CaCl_2` = 65.14 g

Mass of
`Na_2CO_3` = 74.68 g

Molar mass of
`CaCl_2` = 111 g/mol

Molar mass of
`Na_2CO_3` = 106 g/mol

First we have to calculate the moles of
`CaCl_2` and
`Na_2CO_3`.

`\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}`

`\text{Moles of }CaCl_2=(65.14g)/(111g/mol)=0.587mol`

and,

`\text{Moles of }Na_2CO_3=\frac{\text{Given mass }Na_2CO_3}{\text{Molar mass }Na_2CO_3}`

`\text{Moles of }Na_2CO_3=(74.68g)/(106g/mol)=0.704mol`

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

`CaCl_2+Na_2CO_3\rightarrow CaCO_3+2NaCl`

From the balanced reaction we conclude that

As, 1 mole of
`CaCl_2` react with 1 mole of
`Na_2CO_3`

So, 0.587 mole of
`CaCl_2` react with 0.587 mole of
`Na_2CO_3`

From this we conclude that,
`Na_2CO_3` is an excess reagent because the given moles are greater than the required moles and
`CaCl_2` is a limiting reagent and it limits the formation of product.

Hence, the limiting reactant is,
`CaCl_2`