The question is incomplete, here is the complete question:
\(2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2\)
Determine how many grams of KI are required to produce 1.2g of \(PbI_2\)
A) 0.43 g B) 0.86 g C) 1.67 g D) 3.33 g
Answer: The mass of KI needed is 0.86 grams.
Explanation:
To calculate the number of moles, we use the equation:
\(\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}\) .....(1)
Given mass of lead (II) iodide = 1.2 g
Molar mass of lead (II) iodide = 461 g/mol
Putting values in equation 1, we get:
\(\text{Moles of lead (II) iodide}=\frac{1.2g}{461g/mol}=2.60\times 10^{-3}mol\)
The chemical equation for the reaction of potassium iodide
\(2KI+Pb(NO_3)_2\rightarrow 2KNO_3+PbI_2\)
By Stoichiometry of the reaction:
1 mole of lead (II) iodide is produced from 2 moles of KI
So, \(2.60\times 10^{-3}\) moles of lead (II) iodide will be produced from \(\frac{2}{1}\times 2.60\times 10^{-3}=5.2\times 10^{-3}\) moles of KI
Now, calculating the mass of KI by using equation 1, we get:
Molar mass of KI = 166 g/mol
Moles of KI = \(5.2\times 10^{-3}\) moles
Putting values in equation 1, we get:
\(5.2\times 10^{-3}mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=(5.2\times 10^{-3}mol\times 166g/mol)=0.86g\)
Hence, the mass of KI needed is 0.86 grams.
