Question

If 4.00 g of NaCl react with 10.00 g of AgNO3, what is the excess reactant?

Answer 1

The reaction formula of this is NaCl + AgNO3 = NaNO3 + AgCl. The mole number of NaCl is 4/58.5=0.068 mol. The mole number of AgNO3 is 10/170=0.059 mol. So the NaCl is excess.

Answer 2

Answer : The excess reagent is,
`NaCl`

Explanation :

First we have to calculate the moles of
`NaCl` and
`AgNO_3`.

`\text{Moles of }NaCl=\frac{\text{Mass of }NaCl}{\text{Molar mass of }NaCl}=(4g)/(58.4g/mole)=0.068moles`

`\text{Moles of }AgNO_3=\frac{\text{Mass of }AgNO_3}{\text{Molar mass of }AgNO_3}=(10g)/(169.9g/mole)=0.058moles`

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

`NaCl+AgNO_3\rightarrow AgCl+NaNO_3`

From the balanced reaction we conclude that

As, 1 mole of
`NaCl` react with 1 mole of
`AgNO_3`

So, 0.068 moles of
`NaCl` react with 0.068 moles of
`AgNO_3`

From this we conclude that, the moles of
`AgNO_3` are less than the NaCl. So,
`AgNO_3` is a limiting reagent because it limits the formation of products and
`NaCl` is an excess reagent.

Hence, the excess reagent is,
`NaCl`