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
Question 4
Lead nitrate solution reacts with sodium iodide solution to form solid lead iodide and sodium nitrate solution.
(a) (i) Complete the sentence by putting a cross (✗)... show full transcript
Worked Solution & Example Answer: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
Step 1
(i) Complete the sentence by putting a cross (✗) in the box next to your answer.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
This reaction is an example of:
D precipitation (correct answer)
This is because lead iodide forms as a solid from the reaction of two aqueous solutions, illustrating a common precipitation reaction.
Step 2
(ii) Complete the equation by filling in the state symbols for the products.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The balanced equation with state symbols is:
Pb(NO₃)₂(aq) + 2NaI(aq) → PbI₂(s) + 2NaNO₃(aq)
Here, PbI₂ is a solid precipitate, denoted by (s).
Step 3
(iii) Calculate the relative formula mass of sodium nitrate, NaNO₃.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the relative formula mass of sodium nitrate (NaNO₃):
Sodium (Na) = 23
Nitrogen (N) = 14
Oxygen (O) = 16
The formula mass can be calculated as:
extRelativeFormulaMass=23+14+(3imes16)=85
Thus, the relative formula mass of NaNO₃ is 85.
Step 4
(iv) Calculate, using this formula, the percentage by mass of lead in lead iodide.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the percentage by mass of lead in lead iodide (PbI₂):
Relative mass of lead (Pb) = 207
Relative mass of iodine (I) = 127
The relative formula mass of PbI₂ is:
extRelativeFormulaMassofPbI2=207+(2imes127)=461
The percentage by mass of lead in lead iodide can be calculated as:
ext{Percentage by mass} = rac{207}{461} imes 100 \\ \approx 44.87\%
So, the percentage by mass of lead in PbI₂ is approximately 44.87%.
