Question

when 0.56 grams of CuCl2 react with 0.64 grams of Kl, what is the limiting reactant in the production of iodide?

Answer 1

`KI`

Step-by-step explanation:

Here, we want to get the limiting reactant

The limiting reactant is the reactant that produces less amount of the product

We start by writing the balanced equation of the reaction:

We have this as:

`2CuCl_2\text{ + 4KI }\rightarrow\text{ Cu}_2I_2\text{ + I}_2\text{ + 4KCl}`

The iodide here is the copper (i) iodide

From the question, we have 0.56 g of copper (ii) chloride

We need to get the number of moles of it that reacted

We can get that by dividing the mass by the molar mass of copper (ii) chloride

The molar mass of copper (ii) chloride is 134.45 g

Thus, we have the number of moles as:

`(0.56)/(134.45)\text{ = 0.004165 mol}`

From the equation of reaction, 2 moles of copper ii chloride produced 1 mole of the iodide

Thus, the number of moles of the iodide produced by the actual reacting mass would be:

`(0.004165)/(2)\text{ = 0.0020825 mol}`

Now, let us get the number of moles produced by Potassium iodide

We can get this by dividing the mass by the molar mass of potassium iodide

The molar mass of potassium iodide is 166 g/mol

Thus, we have the number of moles as:

`(0.64)/(166)\text{ = 0.00386}`

From the equation of reaction, 4 moles of KI produced 1 mole of the iodide

Thus:

`\begin{gathered} 0.00386\text{ mol will produce:} \\ (0.00386)/(4)\text{ = 0.000965} \end{gathered}`

From what we can see, potassium iodide would give less mass of the copper (i) iodide

This makes potassium iodide the limiting reactant