Question

If 100 grams of potassium (K) reacts with 260 grams of iodine (I2) to form potassium iodide (KI), which is the limiting reactant? If the percent yield is 80% how much KI is produced?

Answer 1

Answer on Question #57246 - Chemistry - Other

Question:

If 100 grams of potassium (K) reacts with 260 grams of iodine (I2) to form potassium iodide (KI), which is the limiting reactant? If the percent yield is 80% how much KI is produced?

Answer:

The reaction equation of potassium and iodine is

$$2 \mathrm {K} + \mathrm {I} 2 \rightarrow 2 \mathrm {K I}$$

Then, one can see that 2 moles of potassium react with 1 mole of iodine. Let's calculate the number of the moles of potassium and iodine given in the task:

$$n (K) = \frac {m (K)}{M (K)} = \frac {100}{39.1} = 2.56 \mathrm{mol}$$ $$n (I _ {2}) = \frac {m (I _ {2})}{M (I _ {2})} = \frac {260}{253.8} = 1.02 \mathrm{mol}$$

According to reaction:

$$n (I _ {2}) = \frac {n (K)}{2}$$

Then, if one compares $2.56 / 2 = 1.28$ and 1.02 mol, one can see that iodine is the limiting reactant.

The percent yield is the ratio of the mass of KI produced to the theoretical value:

$$\eta = \frac {m (K I) - \text{produced}}{m (K I) - \text{theoretical}} \cdot 100\%$$

The theoretical value can be calculated from the limiting reactant quantity:

$$n (I _ {2}) = \frac {n (K I)}{2}, \quad n (K I) = 2 n (I _ {2})$$ $$n (K I) = 2 * 1.02 = 2.04 \mathrm{mol}, m (K I) = n (K I) \cdot M (K I) = 2.04 \cdot 166.0 = 340.1 \mathrm{g}$$

This mass is the theoretical value. Then, we use the percent yield to calculate the mass of potassium iodide produced:

$$m (K I - \exp) = \eta \cdot m (K I - \text{theor}) / 100\% = 0.8 \cdot 340.1 = 272.1 \mathrm{g}$$