how many grams of silver iodide, AgI, are produced from 24.7 g of calcium iodide, CaI2
The correct answer and explanation is :
To determine how many grams of silver iodide (AgI) are produced from 24.7 g of calcium iodide (CaI2), we can use the stoichiometric method, which involves a series of steps: converting grams to moles, using the balanced chemical equation to find the mole ratio, and then converting moles back to grams.
Step 1: Write the balanced chemical equation.
The reaction between calcium iodide (CaI2) and silver nitrate (AgNO3) to form silver iodide (AgI) and calcium nitrate (Ca(NO3)2) is as follows:
[
\text{CaI}_2 (aq) + 2 \text{AgNO}_3 (aq) \rightarrow 2 \text{AgI} (s) + \text{Ca(NO}_3)_2 (aq)
]
This equation shows that 1 mole of CaI2 produces 2 moles of AgI.
Step 2: Calculate the molar mass of calcium iodide (CaI2).
The molar mass of CaI2 is calculated by adding the atomic masses of calcium (Ca) and iodine (I):
- Molar mass of Ca = 40.08 g/mol
- Molar mass of I = 126.90 g/mol
So, the molar mass of CaI2 is:
[
\text{Molar mass of CaI}_2 = 40.08 \, \text{g/mol} + (2 \times 126.90 \, \text{g/mol}) = 293.88 \, \text{g/mol}
]
Step 3: Convert grams of CaI2 to moles.
We have 24.7 g of CaI2, so we can convert this to moles by dividing by the molar mass of CaI2:
[
\text{Moles of CaI}_2 = \frac{24.7 \, \text{g}}{293.88 \, \text{g/mol}} = 0.084 \, \text{mol}
]
Step 4: Use the mole ratio to find moles of AgI.
From the balanced equation, we know that 1 mole of CaI2 produces 2 moles of AgI. Therefore, the moles of AgI produced will be:
[
\text{Moles of AgI} = 0.084 \, \text{mol} \times 2 = 0.168 \, \text{mol}
]
Step 5: Calculate the molar mass of AgI.
The molar mass of AgI is:
- Molar mass of Ag = 107.87 g/mol
- Molar mass of I = 126.90 g/mol
[
\text{Molar mass of AgI} = 107.87 \, \text{g/mol} + 126.90 \, \text{g/mol} = 234.77 \, \text{g/mol}
]
Step 6: Convert moles of AgI to grams.
Now, we can convert moles of AgI to grams:
[
\text{Grams of AgI} = 0.168 \, \text{mol} \times 234.77 \, \text{g/mol} = 39.4 \, \text{g}
]
Final Answer:
Thus, 24.7 g of calcium iodide (CaI2) will produce 39.4 g of silver iodide (AgI).