When lead (II) nitrate reacts with sodium iodide, sodium nitrate and lead (II) iodide are formed. Balance the following equation: Pb(NO3)2 (aq) + 2NaI (aq) → 2NaNO3 (aq) + PbI2 (s) If I start with 25 grams of lead (II) nitrate and 15.0 grams of sodium iodide, how many grams of sodium nitrate can be formed? What is the limiting reagent in the reaction described in problem 2? How many grams of lead (II) iodide is formed? How much of the nonlimiting reagent will be left over from the reaction in problem #2? If 6 grams of sodium nitrate are formed in the reaction described in problem #2, what is the percent yield of this reaction?
The Correct Answer and Explanation is:
Let’s break down the reaction step by step, starting with the balanced equation:
Balanced equation:
Pb(NO3)2(aq)+2NaI(aq)→2NaNO3(aq)+PbI2(s)\text{Pb(NO}_3)_2 (aq) + 2\text{NaI} (aq) \rightarrow 2\text{NaNO}_3 (aq) + \text{PbI}_2 (s)Pb(NO3)2(aq)+2NaI(aq)→2NaNO3(aq)+PbI2(s)
1. Identifying the Limiting Reagent:
We are given 25 grams of lead (II) nitrate (Pb(NO₃)₂) and 15.0 grams of sodium iodide (NaI). To find the limiting reagent, we first need to convert the masses into moles.
Molar Mass of Pb(NO₃)₂:
Pb=207.2 g/mol, N=14.0 g/mol, O=16.0 g/mol\text{Pb} = 207.2 \, \text{g/mol}, \, \text{N} = 14.0 \, \text{g/mol}, \, \text{O} = 16.0 \, \text{g/mol}Pb=207.2g/mol,N=14.0g/mol,O=16.0g/mol
Molar Mass of Pb(NO₃)₂=207.2+2(14.0+3×16.0)=241.2 g/mol\text{Molar Mass of Pb(NO₃)₂} = 207.2 + 2(14.0 + 3 \times 16.0) = 241.2 \, \text{g/mol}Molar Mass of Pb(NO₃)₂=207.2+2(14.0+3×16.0)=241.2g/mol
Moles of Pb(NO₃)₂: Moles of Pb(NO₃)₂=25.0 g241.2 g/mol=0.1036 mol\text{Moles of Pb(NO₃)₂} = \frac{25.0 \, \text{g}}{241.2 \, \text{g/mol}} = 0.1036 \, \text{mol}Moles of Pb(NO₃)₂=241.2g/mol25.0g=0.1036mol
Molar Mass of NaI: Na=22.99 g/mol, I=126.9 g/mol\text{Na} = 22.99 \, \text{g/mol}, \, \text{I} = 126.9 \, \text{g/mol}Na=22.99g/mol,I=126.9g/mol Molar Mass of NaI=22.99+126.9=149.89 g/mol\text{Molar Mass of NaI} = 22.99 + 126.9 = 149.89 \, \text{g/mol}Molar Mass of NaI=22.99+126.9=149.89g/mol
Moles of NaI: Moles of NaI=15.0 g149.89 g/mol=0.1001 mol\text{Moles of NaI} = \frac{15.0 \, \text{g}}{149.89 \, \text{g/mol}} = 0.1001 \, \text{mol}Moles of NaI=149.89g/mol15.0g=0.1001mol
2. Stoichiometry of the Reaction:
From the balanced equation, 1 mole of Pb(NO₃)₂ reacts with 2 moles of NaI. So, for 0.1036 moles of Pb(NO₃)₂, you would need: 0.1036 mol Pb(NO₃)₂×2 mol NaI1 mol Pb(NO₃)₂=0.2072 mol NaI0.1036 \, \text{mol Pb(NO₃)₂} \times \frac{2 \, \text{mol NaI}}{1 \, \text{mol Pb(NO₃)₂}} = 0.2072 \, \text{mol NaI}0.1036mol Pb(NO₃)₂×1mol Pb(NO₃)₂2mol NaI=0.2072mol NaI
However, we only have 0.1001 moles of NaI, which is less than the required 0.2072 moles. Therefore, NaI is the limiting reagent.
3. Amount of Sodium Nitrate Formed:
From the balanced equation, 2 moles of NaI produce 2 moles of NaNO₃. Since NaI is the limiting reagent, we can directly calculate the moles of NaNO₃ produced: Moles of NaNO₃=0.1001 mol NaI×2 mol NaNO₃2 mol NaI=0.1001 mol NaNO₃\text{Moles of NaNO₃} = 0.1001 \, \text{mol NaI} \times \frac{2 \, \text{mol NaNO₃}}{2 \, \text{mol NaI}} = 0.1001 \, \text{mol NaNO₃}Moles of NaNO₃=0.1001mol NaI×2mol NaI2mol NaNO₃=0.1001mol NaNO₃
Molar Mass of NaNO₃: NaNO₃=22.99+14.0+3×16.0=85.0 g/mol\text{NaNO₃} = 22.99 + 14.0 + 3 \times 16.0 = 85.0 \, \text{g/mol}NaNO₃=22.99+14.0+3×16.0=85.0g/mol
Mass of NaNO₃ formed: Mass of NaNO₃=0.1001 mol×85.0 g/mol=8.51 g\text{Mass of NaNO₃} = 0.1001 \, \text{mol} \times 85.0 \, \text{g/mol} = 8.51 \, \text{g}Mass of NaNO₃=0.1001mol×85.0g/mol=8.51g
4. Amount of Lead (II) Iodide Formed:
From the balanced equation, 2 moles of NaI produce 1 mole of PbI₂. So, the moles of PbI₂ formed are: Moles of PbI₂=0.1001 mol NaI×1 mol PbI₂2 mol NaI=0.05005 mol PbI₂\text{Moles of PbI₂} = 0.1001 \, \text{mol NaI} \times \frac{1 \, \text{mol PbI₂}}{2 \, \text{mol NaI}} = 0.05005 \, \text{mol PbI₂}Moles of PbI₂=0.1001mol NaI×2mol NaI1mol PbI₂=0.05005mol PbI₂
Molar Mass of PbI₂: PbI₂=207.2+2×126.9=461.0 g/mol\text{PbI₂} = 207.2 + 2 \times 126.9 = 461.0 \, \text{g/mol}PbI₂=207.2+2×126.9=461.0g/mol
Mass of PbI₂ formed: Mass of PbI₂=0.05005 mol×461.0 g/mol=23.1 g\text{Mass of PbI₂} = 0.05005 \, \text{mol} \times 461.0 \, \text{g/mol} = 23.1 \, \text{g}Mass of PbI₂=0.05005mol×461.0g/mol=23.1g
5. Excess NaI Left Over:
Since NaI is the limiting reagent, there will be no NaI left over because all of it is used up in the reaction.
6. Percent Yield:
We are told that 6 grams of NaNO₃ are formed, but the theoretical yield was calculated as 8.51 grams.
Percent Yield: Percent Yield=Actual YieldTheoretical Yield×100=6.0 g8.51 g×100=70.5%\text{Percent Yield} = \frac{\text{Actual Yield}}{\text{Theoretical Yield}} \times 100 = \frac{6.0 \, \text{g}}{8.51 \, \text{g}} \times 100 = 70.5\%Percent Yield=Theoretical YieldActual Yield×100=8.51g6.0g×100=70.5%
Conclusion:
- The limiting reagent is NaI.
- The mass of sodium nitrate formed is 8.51 grams.
- The mass of lead (II) iodide formed is 23.1 grams.
- NaI will be completely consumed in the reaction.
- The percent yield of sodium nitrate is 70.5%.
