Iodine is prepared both in the laboratory and commercially by adding Cl, (g) to an aqueous solution containing sodium iodide. 2 Nal(aq) + Cl₂(g) → 1₂ (s) + 2 NaCl(aq) How many grams of sodium iodide, Nal, must be used to produce 60.0 g of iodine, I₂? mass:
The Correct Answer and Explanation is :
To calculate the grams of sodium iodide (NaI) needed to produce 60.0 g of iodine (I₂), we’ll use stoichiometry, which involves balancing the chemical equation and converting between grams, moles, and other quantities. Here’s the breakdown:
Step 1: Write the balanced chemical equation.
The given reaction is:
[ 2 \, \text{NaI (aq)} + \text{Cl}_2 \, \text{(g)} \rightarrow \text{I}_2 \, \text{(s)} + 2 \, \text{NaCl (aq)} ]
Step 2: Molar mass of iodine (I₂).
The molar mass of iodine (I₂) is calculated by adding the molar masses of two iodine atoms:
[ 1 \, \text{I} = 126.9 \, \text{g/mol} ]
So:
[ \text{Molar mass of I}_2 = 2 \times 126.9 \, \text{g/mol} = 253.8 \, \text{g/mol} ]
Step 3: Use the mass of iodine to find moles of iodine.
We know the mass of iodine is 60.0 g. To find the number of moles of iodine, we divide the mass by the molar mass of iodine:
[
\text{Moles of I}_2 = \frac{60.0 \, \text{g}}{253.8 \, \text{g/mol}} = 0.236 \, \text{mol of I}_2
]
Step 4: Use the stoichiometry of the reaction.
From the balanced equation, 2 moles of NaI produce 1 mole of I₂. This means that for every mole of iodine produced, 2 moles of sodium iodide are required. Therefore, the moles of NaI required to produce 0.236 moles of iodine are:
[
\text{Moles of NaI} = 0.236 \, \text{mol I}_2 \times \frac{2 \, \text{mol NaI}}{1 \, \text{mol I}_2} = 0.472 \, \text{mol NaI}
]
Step 5: Convert moles of NaI to grams.
The molar mass of NaI is calculated by adding the molar mass of sodium (Na) and iodine (I):
[ \text{Molar mass of NaI} = 22.99 \, \text{g/mol} + 126.9 \, \text{g/mol} = 149.89 \, \text{g/mol} ]
Now, we can calculate the mass of NaI required:
[
\text{Mass of NaI} = 0.472 \, \text{mol NaI} \times 149.89 \, \text{g/mol} = 70.8 \, \text{g NaI}
]
Final Answer:
To produce 60.0 g of iodine (I₂), you need 70.8 grams of sodium iodide (NaI).