Calculate the mass of PbI2 produced by treatment of 0.995 g of KI with excess Pb(NO3)2: a. 1.38 g b. 2.76 g c. 11.1 g d. 1.00 g
The Correct Answer and Explanation is:
To find the mass of PbI₂ produced when 0.995 g of KI reacts with excess Pb(NO₃)₂, follow these steps:
Step 1: Write the balanced chemical equation
2KI+Pb(NO3)2→PbI2+2KNO32KI + Pb(NO_3)_2 \rightarrow PbI_2 + 2KNO_32KI+Pb(NO3)2→PbI2+2KNO3
This tells us that 2 moles of potassium iodide (KI) react with 1 mole of lead(II) nitrate to form 1 mole of lead(II) iodide (PbI₂).
Step 2: Calculate the molar mass of KI
- Potassium (K): 39.10 g/mol
- Iodine (I): 126.90 g/mol
- Molar mass of KI = 39.10 + 126.90 = 166.00 g/mol
Now, calculate moles of KI:Moles of KI=0.995 g166.00 g/mol=0.00599 mol\text{Moles of KI} = \frac{0.995\ \text{g}}{166.00\ \text{g/mol}} = 0.00599\ \text{mol}Moles of KI=166.00 g/mol0.995 g=0.00599 mol
Step 3: Use stoichiometry to find moles of PbI₂ formed
From the balanced equation:2 mol KI→1 mol PbI22\ \text{mol KI} \rightarrow 1\ \text{mol PbI}_22 mol KI→1 mol PbI2Moles of PbI2=0.005992=0.002995 mol\text{Moles of PbI}_2 = \frac{0.00599}{2} = 0.002995\ \text{mol}Moles of PbI2=20.00599=0.002995 mol
Step 4: Calculate the mass of PbI₂
- Lead (Pb): 207.2 g/mol
- Iodine (I₂): 2 × 126.9 = 253.8 g/mol
- Molar mass of PbI₂ = 207.2 + 253.8 = 461.0 g/mol
Mass of PbI2=0.002995 mol×461.0 g/mol=1.38 g\text{Mass of PbI}_2 = 0.002995\ \text{mol} \times 461.0\ \text{g/mol} = 1.38\ \text{g}Mass of PbI2=0.002995 mol×461.0 g/mol=1.38 g
Correct answer: a. 1.38 g
Explanation:
In this reaction, KI is the limiting reagent, and Pb(NO₃)₂ is in excess. We used the balanced equation to determine the mole ratio. Then we converted grams of KI to moles, applied the stoichiometry to find the moles of PbI₂, and converted that to grams. This approach ensures the correct theoretical yield is calculated based on limiting reactant principles and molar mass relationships.
