Question

determine how many grams of KI are required to produce 1.2g of PbI2. A) 0.43 g B) 0.86 g C) 1.67 g D) 3.33 g

Answer 1

2KI + Pb(NO3)2 → 2KNO3 + PbI2 so B 0.86

Answer 2

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.

Step-by-step 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}=(1.2g)/(461g/mol)=2.60* 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* 10^(-3)` moles of lead (II) iodide will be produced from
`(2)/(1)* 2.60* 10^(-3)=5.2* 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* 10^(-3)` moles

Putting values in equation 1, we get:

`5.2* 10^(-3)mol=\frac{\text{Mass of KI}}{166g/mol}\\\\\text{Mass of KI}=(5.2* 10^(-3)mol* 166g/mol)=0.86g`

Hence, the mass of KI needed is 0.86 grams.