What mass of sodium chloride is produced when chlorine reacts with 0.29 g of sodium iodide?
The Correct Answer and Explanation is :
To determine the mass of sodium chloride (NaCl) produced when chlorine reacts with sodium iodide (NaI), we need to understand the reaction and stoichiometry involved.
The Reaction:
When chlorine (Cl₂) reacts with sodium iodide (NaI), a halogen displacement reaction occurs:
[
Cl_2 (g) + 2NaI (aq) \rightarrow 2NaCl (aq) + I_2 (s)
]
In this reaction, chlorine displaces iodine from sodium iodide, resulting in the formation of sodium chloride (NaCl) and iodine (I₂).
Step 1: Molar Mass Calculation
To solve this problem, we need to convert the mass of sodium iodide (NaI) to moles, use stoichiometry to find the moles of NaCl produced, and then convert the moles of NaCl back to mass.
- Molar mass of NaI:
- Sodium (Na): 22.99 g/mol
- Iodine (I): 126.90 g/mol
- Molar mass of NaI = 22.99 g/mol + 126.90 g/mol = 149.89 g/mol
- Molar mass of NaCl:
- Sodium (Na): 22.99 g/mol
- Chlorine (Cl): 35.45 g/mol
- Molar mass of NaCl = 22.99 g/mol + 35.45 g/mol = 58.44 g/mol
Step 2: Calculate Moles of Sodium Iodide
Given that the mass of sodium iodide is 0.29 g, we calculate the number of moles of NaI:
[
\text{moles of NaI} = \frac{\text{mass of NaI}}{\text{molar mass of NaI}} = \frac{0.29 \, \text{g}}{149.89 \, \text{g/mol}} = 0.00194 \, \text{mol}
]
Step 3: Stoichiometric Relationship
From the balanced equation, we know that 1 mole of Cl₂ reacts with 2 moles of NaI to produce 2 moles of NaCl. Therefore, the moles of NaCl produced are equal to the moles of NaI used (since the ratio of NaI to NaCl is 1:1):
[
\text{moles of NaCl} = 0.00194 \, \text{mol}
]
Step 4: Calculate the Mass of Sodium Chloride
Now, using the moles of NaCl and its molar mass, we can calculate the mass of NaCl produced:
[
\text{mass of NaCl} = \text{moles of NaCl} \times \text{molar mass of NaCl} = 0.00194 \, \text{mol} \times 58.44 \, \text{g/mol} = 0.113 \, \text{g}
]
Conclusion:
The mass of sodium chloride produced when 0.29 g of sodium iodide reacts with chlorine is 0.113 g.