Question

What is the limiting reagent in the reaction?How many grams of lead(II) iodide is formed?How much of the excess reagent will be left over from the reaction?If 6 grams of sodium nitrate are formed in the reaction what is the percent yield of this reaction? 25.0 grams of lead (II) nitrate and 15.0 grams of sodium iodide1Pb(NO3)2 (aq) + 2NaL(aq) → 1PbL2 (s) + 2NaNO3 (aq)

Answer 1

1. NaI is the limiting reactant,

2. mass of PbI2 = 92.202 g

3. Excess = 4.25 g

4. %Yield = 70.6%

Step-by-step explanation

Given

1Pb(NO3)2 (aq) + 2NaI(aq) → 1PbI2 (s) + 2NaNO3 (aq)

mass of Pb(NO3)2 = 25.0 g

mass of NaI = 15.0 g

We know the molar masses:

Molar mass of Pb(NO3)2 = 331.2 g/mol

Molar mass of NaI = 149.89 g/mol

Molar mass of NaNO3 = 84.9947 g/mol

Molar mass of PbI2 = 461.01 g/mol

Required:

1. Limiting reagent in the reaction.

2. Grams of lead(II) iodide is formed.

3. Excess reagent that will be left over from the reaction.

4. If 6 grams of sodium nitrate are formed in the reaction what is the percent yield of this reaction?

Solutions:

1. Limiting reagent

For Pb(NO3)2

`\begin{gathered} 25.0\text{ g Pb\lparen NO}_3)_2\text{ x }\frac{1\text{ mol Pb\lparen NO}_3)_2}{331.2g\text{ Pb\lparen NO}_3)_2}\text{ x }\frac{2\text{ mol NaNO}_3}{1\text{ mol Pb\lparen NO}_3)_2}\text{ } \\ \\ =\text{ 0.15 moles of NaNO}_3 \end{gathered}`

For NaI

`\begin{gathered} 15\text{ g NaI x }\frac{1\text{ mol NaI}}{149.89\text{ g NaI}}\text{ x }\frac{2\text{ mol NaNO}_3}{2\text{ mol NaI}} \\ \\ =\text{ 0.10 mol NaNO}_3 \end{gathered}`

Therefore NaI is the limiting reactant, since it will produce less moles of NaNO3.

2. Mass of PbI2 that will be formed.

Use the limiting reactant to solve this problem.

Step 1: Find the moles of PbI2

The molar ratio between NaI and PbI2 is 2:1

The number of moles of PbI2 = (15g/149.89g.mol^-1) x (1/2) = 0.20 mol of PbI2

Step 2: Convert moles to mass of PbI2

m = n x M

m = 0.20 mol x 461.01 g/mol

m = 92.202 g

3. Excess reagent that will be left over from the reaction.

We know that an excess reactant is Pb(NO3)2

From 1., we know that:-

For Pb(NO3)2 = 0.15 moles of NaNO3 will be produced

For NaI = 0.10 moles of NaNO3 will be produced.

To know the excess, first find the mass of NaNO3 will be produced from both Pb(NO3)2 and NaI.

For Pb(NO3)2

m = 0.15 mol x 84.9947 g/mol = 12.75 g of NaNO3

For NaI

m = 0.10 mol x 84.9947 g/mol = 8.5 g of NaNO3

Excess = 12.75 - 8.5 = 4.25 g

4. If 6 grams of sodium nitrate are formed in the reaction what is the percent yield of this reaction?

%Yield = (actual yield/theoretical yield) x 100

%Yield = (6 g/8.5 g) x 100

%Yield = 70.6%