7.04 g of lead (II) nitrate Pb(NO3)2 has how many moles of the nitrate ion?
The correct answer and explanation is:
Correct Answer: 0.0426 moles of nitrate ion (NO₃⁻)
To determine how many moles of nitrate ion are present in 7.04 grams of lead(II) nitrate (Pb(NO₃)₂), follow these steps:
Step 1: Determine the molar mass of Pb(NO₃)₂.
Break down the molar masses of each element:
- Pb = 207.2 g/mol
- N = 14.01 g/mol
- O = 16.00 g/mol
There are two nitrate ions in one molecule of Pb(NO₃)₂. Each nitrate (NO₃⁻) has:
- 1 N = 14.01 g/mol
- 3 O = 3 × 16.00 = 48.00 g/mol
- Total for one NO₃⁻ = 14.01 + 48.00 = 62.01 g/mol
So, for Pb(NO₃)₂:
- Pb = 207.2 g/mol
- 2 × NO₃⁻ = 2 × 62.01 = 124.02 g/mol
- Total molar mass = 207.2 + 124.02 = 331.22 g/mol
Step 2: Convert grams of Pb(NO₃)₂ to moles. Moles of Pb(NO₃)₂=7.04 g331.22 g/mol≈0.02125 mol\text{Moles of Pb(NO₃)₂} = \frac{7.04\ \text{g}}{331.22\ \text{g/mol}} \approx 0.02125\ \text{mol}
Step 3: Determine moles of nitrate ions.
Each mole of Pb(NO₃)₂ gives 2 moles of NO₃⁻ ions. Moles of NO₃⁻=0.02125 mol Pb(NO₃)₂×2=0.0425 mol\text{Moles of NO₃⁻} = 0.02125\ \text{mol Pb(NO₃)₂} × 2 = \mathbf{0.0425\ mol}
Rounded to 3 significant figures: Final Answer=0.0426 mol of NO₃⁻\text{Final Answer} = \boxed{0.0426\ \text{mol of NO₃⁻}}
Explanation:
The question tests your ability to apply stoichiometry and molar mass concepts. Lead(II) nitrate is an ionic compound made up of one Pb²⁺ ion and two NO₃⁻ ions. When dissolved or analyzed chemically, each mole of lead(II) nitrate yields two moles of nitrate ions. This means we multiply the moles of the compound by two to get the moles of nitrate. Converting grams to moles is done by dividing the given mass by the molar mass. This is a basic and essential skill in chemistry for quantifying ions or molecules in a sample.