Lead (IV) sulfate reacts with lithium nitrate, write a balanced equation to represent this reaction.
How many grams of lithium nitrate will be needed to make 250. grams of lithium sulfate. assuming that you have an adequate amount of lead (IV) sulfate?
The Correct Answer and Explanation is :
To start, let’s write the balanced chemical equation for the reaction between lead (IV) sulfate and lithium nitrate to form lithium sulfate and lead (IV) nitrate.
The reaction is:
[
\text{Pb(SO}_4\text{)}_2 + 4 \text{LiNO}_3 \rightarrow 2 \text{Li}_2\text{SO}_4 + \text{Pb(NO}_3\text{)}_4
]
In this equation, lead (IV) sulfate reacts with lithium nitrate to form lithium sulfate and lead (IV) nitrate.
Step 1: Determining the Molar Masses
To find the mass of lithium nitrate required to produce 250 g of lithium sulfate, we need to follow a few steps, beginning with determining the molar masses.
- Molar Mass of Lithium Sulfate (Li₂SO₄):
- Lithium (Li): 6.94 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol
- Molar mass of Li₂SO₄ = (2 × 6.94) + 32.07 + (4 × 16.00) = 45.88 + 32.07 + 64.00 = 141.95 g/mol
- Molar Mass of Lithium Nitrate (LiNO₃):
- Lithium (Li): 6.94 g/mol
- Nitrogen (N): 14.01 g/mol
- Oxygen (O): 16.00 g/mol
- Molar mass of LiNO₃ = 6.94 + 14.01 + (3 × 16.00) = 6.94 + 14.01 + 48.00 = 68.95 g/mol
Step 2: Using Stoichiometry to Calculate Required Mass of Lithium Nitrate
From the balanced equation, we know that 1 mole of Pb(SO₄)₂ produces 2 moles of Li₂SO₄. For every 1 mole of Li₂SO₄ produced, 4 moles of LiNO₃ are required.
We can use this information to set up a stoichiometric calculation.
- Determine moles of Li₂SO₄ needed:
- Mass of Li₂SO₄ required = 250.0 g
- Moles of Li₂SO₄ = mass / molar mass = 250.0 g / 141.95 g/mol = 1.76 moles of Li₂SO₄
- Calculate moles of LiNO₃ required:
- From the balanced equation, 4 moles of LiNO₃ are required for every 2 moles of Li₂SO₄. Therefore, moles of LiNO₃ = (1.76 moles Li₂SO₄) × (4 moles LiNO₃ / 2 moles Li₂SO₄) = 3.52 moles of LiNO₃
- Convert moles of LiNO₃ to grams:
- Mass of LiNO₃ = moles × molar mass = 3.52 moles × 68.95 g/mol = 242.6 g
Conclusion:
To make 250.0 grams of lithium sulfate, 242.6 grams of lithium nitrate will be needed, assuming an adequate amount of lead (IV) sulfate is available.