Question: Given the reaction \[Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} + 2KI \to Pb{I_2}_{\left( ...

Question

Given the reaction
$Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} + 2KI \to Pb{I_2}_{\left( s \right)} + 2KN{O_{3\left( {aq} \right)}}$
What is the mass of $Pb{I_2}$ that will precipitate if $10.2$$$$Pb{\left( {N{O_3}} \right)_2}$ is mixed with $5.73$ g of $KI$ in a sufficient quantity of water?
A. $2.06g$
B. $4.13g$
C. $7.96g$
D. $15.9g$

Answer 1

Solution

Here one mole of lead nitrate is reacting with two potassium iodide and there is a formation of one mole of lead iodide with two mole potassium nitrate. The lead nitrate is an inorganic compound having the chemical formula $Pb{\left( {N{O_3}} \right)_2}$ which is a colorless crystal or it occurs as a white powder. The potassium iodide is a stable iodide salt with formula $KI$ .

Complete answer:
The mass of lead iodide is not equal to $2.06g$ . Hence, option (A) is incorrect.
By substituting the given values in the corresponding equation, we will not get the value of mass of iodide as $4.13g$ . Hence the option (B) is incorrect.
According to the question, if lead nitrate is mixed with potassium iodide, there will be lead iodide with potassium nitrate. The reaction is,
$Pb{\left( {N{O_3}} \right)_{2\left( {aq} \right)}} + 2KI \to Pb{I_2}_{\left( s \right)} + 2KN{O_{3\left( {aq} \right)}}$
Here, the number of moles of lead nitrate is equal to one and the molecular weight of lead nitrate is $331.2g$ .
Number of moles of potassium iodide is equal to two and its molecular weight is $166g$ .
One of lead nitrates reacts with two mol of potassium iodide which gives one mole of lead iodide.
The molecular weight of lead iodide is equal to $461g$ . Here we have to find out the required mass of $Pb{I_2}$ that will precipitate if $10.2$ $Pb{\left( {N{O_3}} \right)_2}$ is mixed with $5.73$ g of KI in a sufficient quantity of water.
Given mass of lead nitrate is equal to $10.2$ g and potassium iodide is equal to $5.73$ g. Hence, the potassium iodide acts as the limiting reagent. Therefore, it will determine the product.
$5.73$ g of $KI$ gives $= \dfrac{{461}}{{332}} \times 5.73$
$= 7.96g$

So, the correct answer is “Option C”.

Note:
As we know, the limiting reagent is also known as limiting reactant. It is the reactant in a chemical reaction which is completely consumed in a chemical reaction is known as a limiting reagent. Hence, the amount of product produced in the chemical reaction is limited by this reagent. Therefore, after the completion of the limiting reagent, the reaction will not be continued.