Question

249 g of potassium iodide, KI, is mixed with 496.5 g of lead(II) nitrate, Pb(NO3)2.

The equation of the reaction is represented below.



[Pb(NO3)2 = 331; KI = 166; PbI2 = 461, KNO3 = 101]

Calculate the number of moles of excess reagent left.

Give your answer to three significant figures.

Answer 1

Answer: The number of moles of excess reagent left is 0.750 moles

Explanation:

The number of moles is calculated by using the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

• For KI:

Given mass of $KI$ = 249 g

Molar mass of $KI$ = 166 g/mol

Putting values in equation 1, we get:

$\text{Moles of KI}=\frac{249g}{166g/mol}=1.5mol$

• For $Pb(NO_3)_2$:

Given mass of $Pb(NO_3)_2$ = 496.5 g

Molar mass of $Pb(NO_3)_2$ = 331 g/mol

Putting values in equation 1, we get:

$\text{Moles of }Pb(NO_3)_2=\frac{496.5g}{331g/mol}=1.5mol$

The given chemical equation follows:

$2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2$

By stoichiometry of the reaction:

If 2 moles of KI reacts with 1 mole of lead(II) nitrate

So, 1.5 moles of KI will react with = $\frac{1}{2}\times 1.5=0.75mol$ of lead(II) nitrate

As the given amount of lead(II) nitrate is more than the required amount. Thus, it is present in excess and is considered as an excess reagent.

Thus, KI is considered a limiting reagent because it limits the formation of the product.

Moles of excess reactant ($Pb(NO_3)_2$) left = [1.5 - 0.75] = 0.750 moles

Hence, the number of moles of excess reagent left is 0.750 moles