What mass of iodine is produced when 7.1 g of chlorine reacts with potassium iodide?

Question

Wyatt Atkins · Accepted Answer

We need a stoichiometric equation.............and predict that approx. #25*g# of #I_2# will result.............

#Cl_2(aq) + 2KI(aq) rarr I_2(s) + 2KCl(aq)# It is important to remember the balanced equation, which states that one equiv of diiodine is produced from one equiv of dichlorine. You will need to learn that fluorine is a stronger oxidant than chlorine (in fact, it is the strongest oxidant ever), that bromine is a stronger oxidant than iodine, and that chlorine is a stronger oxidant than fluorine. Because the valence electron of the lower halogen has a higher valence shell than the upper halogen, it tends to be donated to the halogen above it, reducing the upper halogen and OXIDIZING the starting halogen. This is why the halogen lower on the table is a stronger oxidant. We could use the redox half equation to demonstrate this: #1/2Cl_2(aq) + e^(-) rarr Cl^(-)(aq)# The iodide anion is, of course, the reductant and source of electrons: #I^(-) rarr 1/2I_2(s) + e^-# To finally get to your question, we have a molar quantity of #(7.1*g)/(70.9*g*mol^-1)# #=# #0.100*mol# of dichlorine gas. And given the equation, we get a stoichiometric quantity of #I_2#, i.e. #2xx126.9*g*mol^-1xx0.100*mol~=25*g#

Jeremiah Adams · Answer

undefined To find the mass of iodine produced when 7.1 g of chlorine reacts with potassium iodide, we need to first write the balanced chemical equation for the reaction between chlorine and potassium iodide.

The balanced equation is:

Cl2 (g) + 2 KI (aq) → 2 KCl (aq) + I2 (s)

From the balanced equation, we can see that 1 mole of chlorine (Cl2) reacts with 2 moles of potassium iodide (KI) to produce 1 mole of iodine (I2).

Next, we need to calculate the number of moles of chlorine present in 7.1 g of chlorine using its molar mass, which is approximately 35.45 g/mol.

Number of moles of Cl2 = Mass of Cl2 / Molar mass of Cl2
                      = 7.1 g / 35.45 g/mol
                      ≈ 0.2 moles of Cl2

Since the ratio of Cl2 to I2 in the balanced equation is 1:1, the number of moles of iodin

HIX Tutor · Answer

undefined To find the mass of iodine produced, follow these steps:

1. **Write the reaction**: 
   The reaction of chlorine (Cl₂) with potassium iodide (KI) produces potassium chloride (KCl) and iodine (I₂).
   $$
   Cl_2 + 2KI → 2KCl + I_2
   $$

2. **Find the molar mass**:
   - Molar mass of Cl = 35.5 g/mol
   - Molar mass of I = 126.9 g/mol

3. **Calculate moles of chlorine**:
   - 7.1 g Cl₂ has 2 Cl atoms, so:
   $$
   \text{Molar mass of Cl}_2 = 2 \times 35.5 = 71 \text{ g/mol}
   $$
   - Moles of Cl₂:
   $$
   \text{Moles of Cl}_2 = \frac{7.1 \text{ g}}{71 \text{ g/mol}} \approx 0.1 \text{ mol}
   $$

4. **Use the stoichiometry**:
   From the balanced equation, 1 mol of Cl₂ gives 1 mol of I₂.
   $$
   \text{Moles of I}_2 = 0.1 \text{ mol}
   $$

5. **Calculate mass of I₂ produced**:
   $$
   \text{Mass of I}_2 = \text{Moles} \times \text{Molar mass} = 0.1 \text{ mol} \times 253.8 \text{ g/mol} = 25.38 \text{ g}
   $$

6. **Round the answer**:
   The mass of iodine produced is about 25.4 g.

So, **25.4 g of iodine** is produced when 7.1 g of chlorine reacts with potassium iodide.