In aqueous solution, sodium iodide reacts with lead(II) nitrate to give solid lead(II) iodide and aqueous sodium nitrate, as shown in this equation. In one experiment, it is found that 8.34 grams of PbI2 were produced.
2 Nal (aq) + Pb(NO3)2 (aq) → Pbl₂ (s) + 2 NaNO3 (aq)
Based on this reaction,
a. How many moles of PbI2 were produced?
b. How many moles of NaNO3 were produced?
c. How many moles of Nal were consumed?
d. How many grams of Nal were consumed?
The Correct Answer and Explanation is :
To solve this problem, we’ll follow a step-by-step approach using the stoichiometry of the reaction:
Balanced Chemical Equation
[ 2 \text{NaI (aq)} + \text{Pb(NO}_3\text{)}_2 \text{ (aq)} \rightarrow \text{PbI}_2 \text{ (s)} + 2 \text{NaNO}_3 \text{ (aq)} ]
a. Moles of PbI2 Produced
To find the moles of (\text{PbI}_2) produced, use its molar mass. The molar mass of (\text{PbI}_2) (lead(II) iodide) is:
- Lead (Pb): (207.2 \, \text{g/mol})
- Iodine (I): (126.9 \, \text{g/mol}) (each, and there are two iodines)
[ \text{Molar mass of PbI}_2 = 207.2 + 2 \times 126.9 = 460.0 \, \text{g/mol} ]
Given that (8.34 \, \text{grams}) of (\text{PbI}_2) were produced:
[ \text{Moles of PbI}_2 = \frac{8.34 \, \text{g}}{460.0 \, \text{g/mol}} = 0.01813 \, \text{moles} ]
b. Moles of NaNO3 Produced
According to the balanced chemical equation, for every mole of (\text{PbI}_2) produced, 2 moles of (\text{NaNO}_3) are produced. Thus:
[ \text{Moles of NaNO}_3 = 2 \times 0.01813 \, \text{moles} = 0.03626 \, \text{moles} ]
c. Moles of NaI Consumed
The equation shows that 2 moles of (\text{NaI}) are consumed for every mole of (\text{PbI}_2) formed. Therefore:
[ \text{Moles of NaI} = 2 \times 0.01813 \, \text{moles} = 0.03626 \, \text{moles} ]
d. Grams of NaI Consumed
The molar mass of (\text{NaI}) is:
- Sodium (Na): (22.99 \, \text{g/mol})
- Iodine (I): (126.9 \, \text{g/mol})
[ \text{Molar mass of NaI} = 22.99 + 126.9 = 149.89 \, \text{g/mol} ]
Using the moles of (\text{NaI}) consumed:
[ \text{Grams of NaI} = 0.03626 \, \text{moles} \times 149.89 \, \text{g/mol} = 5.434 \, \text{grams} ]
Explanation
In chemical reactions, stoichiometry is used to determine the quantitative relationships between reactants and products. In this specific reaction, sodium iodide reacts with lead(II) nitrate in a double displacement reaction, where ions are exchanged to form a new product, solid lead(II) iodide, and sodium nitrate remains in solution. By knowing the amount of one product, we can determine the amounts of other substances involved in the reaction using the coefficients in the balanced equation, which represent the mole ratios of reactants and products. This process highlights the importance of balanced chemical equations in predicting the outcomes of chemical reactions and ensuring that calculations regarding the amounts of substances involved are accurate.
Image Generation
To visualize this concept, let’s generate an image depicting the reaction setup with labeled reactants and products.
Prompt: A detailed scientific illustration of a laboratory setup for a chemical reaction between sodium iodide and lead(II) nitrate resulting in solid lead(II) iodide and aqueous sodium nitrate. The setup includes beakers, a balance scale for weighing chemicals, safety equipment like gloves and goggles, and labeled containers for each chemical. The image should also include annotations for the reaction equation and stoichiometric calculations. The style should be clean and educational, suitable for a chemistry textbook.
I’ll generate this image now.
Here’s the illustration depicting the laboratory setup for the reaction between sodium iodide and lead(II) nitrate. This setup includes the necessary equipment, safety gear, and clearly labeled chemicals, along with annotations for the reaction equation and stoichiometric calculations. This image is designed to be educational, ideal for understanding the components and the flow of the chemical reaction. You can view the details in the image above.
