Calculate the volume of a sample of iron that has a density of 6.8 g/mL and a mass of 6780 mg.

An aqueous solution containing 7.96 g7.96 g of lead(II) nitrate is added to an aqueous solution containing 6.82 g6.82 g of potas

Eduardwww [97]

Answer: Lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

Explanation:

To calculate the number of moles, we use the equation

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

For lead nitrate:

Given mass of lead nitrate = 7.96 g

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

$\text{Moles of lead nitrate}=\frac{7.96g}{331.2g/mol}=0.024mol$

For potassium chloride:

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

$\text{Moles of potassium chloride}=\frac{6.82g}{74.55g/mol}=0.091mol$

For the given chemical equation:

$Pb(NO_3)_2(aq.)+2KCl(aq.)\rightarrow PbCl_2(s)+2KNO_3(aq.)$

By Stoichiometry of the reaction:

1 mole of lead nitrate reacts with 2 moles of potassium chloride.

So, 0.024 moles of lead nitrate will react with = $\frac{1}{2}\times 0.024=0.048moles$ of potassium chloride

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

Thus, lead nitrate is considered as a limiting reagent because it limits the formation of product.

Amount of excess reagent (potassium chloride) left = 0.091 - 0.048 = 0.043 moles

Now, calculating the mass of potassium chloride from equation 1, we get:

Given mass of potassium chloride = 6.82 g

Molar mass of potassium chloride = 74.55 g/mol

Putting values in above equation, we get:

$0.043mol=\frac{\text{Mass of potassium chloride}}{74.55g/mol}\\\\\text{Mass of potassium chloride}=3.205g$

By Stoichiometry of the reaction:

1 mole of lead nitrate produces 1 mole of lead chloride.

So, 0.024 moles of lead nitrate will produce = $\frac{1}{1}\times 0.024=0.024moles$ of lead chloride

Now, calculating the mass of lead chloride from equation 1, we get:

Molar mass of lead chloride = 278.1 g/mol

Moles of lead chloride = 0.024 moles

Putting values in equation 1, we get:

$0.024mol=\frac{\text{Mass of lead chloride}}{278.1g/mol}\\\\\text{Mass of lead chloride}=6.6744g$

We are given:

Percentage yield of the reaction = 89.3 %

So, amount of lead chloride recovered will be = $\frac{89.3}{100}\times 6.6744g=5.96g$

Hence, lead nitrate is the limiting reagent and potassium chloride is the excess reagent. the amount of excess reagent left is 3.205 grams. The amount of precipitate (lead chloride) recovered is 5.96 g.

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