Lead nitrate solution reacts with sodium iodide solution to form solid lead iodide and sodium nitrate solution - Edexcel - GCSE Chemistry - Question 4 - 2017 - Paper 1

The balanced equation for the reaction is:

$$Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)$$

In this equation, PbI₂ is the solid precipitate formed during the reaction, represented by (s) for solid.

To calculate the relative formula mass for NaNO₃, we need to sum the atomic masses of all the elements in the formula:

• Sodium (Na) = 23
• Nitrogen (N) = 14
• Oxygen (O) = 16 (there are 3 oxygen atoms)

So the calculation will be:

$$ext{Relative formula mass} = 23 + 14 + (3 imes 16) = 23 + 14 + 48 = 85$$

Thus, the relative formula mass of NaNO₃ is 85.

To find the percentage by mass of lead in PbI₂, we first need to calculate the mass of lead and the total mass of PbI₂:

• Lead (Pb) = 207
• Iodine (I) = 127 (there are 2 iodine atoms)

Total mass of PbI₂:

$$ext{Relative formula mass of PbI₂} = 207 + (2 imes 127) = 207 + 254 = 461$$

Now, we can calculate the percentage by mass of lead:

ext{Percentage by mass of lead} = rac{207}{461} imes 100 \\ ext{Percentage by mass of lead} eq 44.9\ ext{ (rounded to one decimal place)}