To find out how many grams of lithium nitrate (LiNO₃) are needed to produce 125.0 grams of lead (IV) nitrate [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex] assuming that there is an adequate amount of lead (IV) sulfate [tex]\( \text{Pb(SO}_4\text{)}_2 \)[/tex], we need to follow a detailed, step-by-step approach using stoichiometry.
### Step 1: Molar Mass Calculation
First, let’s revise the molar masses of the compounds involved in the reaction:
- Molar mass of [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex] = 455.24 g/mol
- Molar mass of [tex]\( \text{LiNO}_3 \)[/tex] = 68.94 g/mol
### Step 2: Convert Mass of [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex] to Moles
Now, convert the given mass of [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex] to moles:
[tex]\[ \text{Moles of }\text{Pb(NO}_3\text{)}_4 = \frac{\text{Mass of Pb(NO}_3\text{)}_4}{\text{Molar Mass of Pb(NO}_3\text{)}_4} = \frac{125.0 \text{ g}}{455.24 \text{ g/mol}} \approx 0.2746 \text{ mol} \][/tex]
### Step 3: Stoichiometric Calculation to Find Moles of LiNO₃ Needed
According to the balanced chemical equation, 1 mole of [tex]\( \text{Pb(SO}_4\text{)}_2 \)[/tex] reacts with 4 moles of [tex]\( \text{LiNO}_3 \)[/tex] to produce 1 mole of [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex].
Therefore, for the amount of [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex] given (0.2746 mol), we can use the stoichiometric ratio (1:4) to find the moles of [tex]\( \text{LiNO}_3 \)[/tex] required:
[tex]\[ \text{Moles of LiNO}_3 = 4 \times \text{Moles of Pb(NO}_3\text{)}_4 \approx 4 \times 0.2746 = 1.0983 \text{ mol} \][/tex]
### Step 4: Convert Moles of LiNO₃ to Mass
Now, convert the moles of LiNO₃ to grams using its molar mass:
[tex]\[ \text{Mass of LiNO}_3 = \text{Moles of LiNO}_3 \times \text{Molar Mass of LiNO}_3 = 1.0983 \text{ mol} \times 68.94 \text{ g/mol} \approx 75.72 \text{ g} \][/tex]
### Conclusion
You will need approximately 75.72 grams of lithium nitrate [tex]\( \text{LiNO}_3 \)[/tex] to produce 125.0 grams of lead (IV) nitrate [tex]\( \text{Pb(NO}_3\text{)}_4 \)[/tex], assuming an adequate amount of lead (IV) sulfate [tex]\( \text{Pb(SO}_4\text{)}_2 \)[/tex].
