Question

+

For the reaction Pb(NO3)2 + 2KI Pbl2 + 2KNO3, how many moles of lead iodide are
produced from 237.1 g of potassium iodide?
Select one:
O a. 0.714
Ob. 5.47e4
O c. 1.54
Od 1.54

Answer 1

The correct awnser is A- 0.174

Answer 2

Answer: 0.714 moles of
`PbI_2` will be produced from 237.1 g of potassium iodide

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of potassium iodide}=(237.1g)/(166g/mol)=1.428moles`

The balanced chemical reaction is:

`Pb(NO_3)_2+2KI\rightarrow PbI_2+2KNO_3`

According to stoichiometry :

2 moles of
`KI` produce = 1 mole of
`PbI_2`

Thus moles of
`KI` will require=
`(1)/(2)* 1.428=0.714moles` of
`PbI_2`

Thus 0.714 moles of
`PbI_2` will be produced from 237.1 g of potassium iodide