Photo AI
An experiment involves reacting 0.02 moles of silver ions with ions of a group 7 element to form 2.868 g of precipitate - Scottish Highers Chemistry - Question 24 - 2022
Question 24
An experiment involves reacting 0.02 moles of silver ions with ions of a group 7 element to form 2.868 g of precipitate. Which of the following is the precipitate? ... show full transcript
Worked Solution & Example Answer:An experiment involves reacting 0.02 moles of silver ions with ions of a group 7 element to form 2.868 g of precipitate - Scottish Highers Chemistry - Question 24 - 2022
Step 1
Determine the molar mass of the precipitate
Answer
To find which silver halide is the precipitate, we first need to determine the molar mass of the potential precipitates. We will also use the mass of the precipitate (2.868 g) to find the moles formed.
Calculating the moles of the precipitate:
[ \text{Moles} = \frac{\text{mass}}{\text{molar mass}} ]
Since we will estimate the molar masses:
- Molar mass of Silver(I) chloride (AgCl) = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol
- Molar mass of Silver(I) bromide (AgBr) = 107.87 (Ag) + 79.90 (Br) = 187.77 g/mol
- Molar mass of Silver(I) iodide (AgI) = 107.87 (Ag) + 126.90 (I) = 234.77 g/mol
- Molar mass of Silver(I) fluoride (AgF) = 107.87 (Ag) + 19.00 (F) = 126.87 g/mol
Step 2
Calculate the moles of the precipitate formed
Answer
Using the molar mass of Silver(I) chloride as an example:
[ \text{Moles of AgCl} = \frac{2.868 \text{ g}}{143.32 \text{ g/mol}} \approx 0.0200 \text{ moles} ]
For Silver(I) bromide:
[ \text{Moles of AgBr} = \frac{2.868 \text{ g}}{187.77 \text{ g/mol}} \approx 0.0153 \text{ moles} ]
For Silver(I) iodide:
[ \text{Moles of AgI} = \frac{2.868 \text{ g}}{234.77 \text{ g/mol}} \approx 0.0122 \text{ moles} ]
For Silver(I) fluoride:
[ \text{Moles of AgF} = \frac{2.868 \text{ g}}{126.87 \text{ g/mol}} \approx 0.0226 \text{ moles} ]
Step 3