Answer: 0.714 moles of [tex]PbI_2[/tex] will be produced from 237.1 g of potassium iodide
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of potassium iodide}=\frac{237.1g}{166g/mol}=1.428moles[/tex]
The balanced chemical reaction is:
[tex]Pb(NO_3)_2+2KI\rightarrow PbI_2+2KNO_3[/tex]
According to stoichiometry :
2 moles of [tex]KI[/tex] produce = 1 mole of [tex]PbI_2[/tex]
Thus moles of [tex]KI[/tex] will require=[tex]\frac{1}{2}\times 1.428=0.714moles[/tex] of [tex]PbI_2[/tex]
Thus 0.714 moles of [tex]PbI_2[/tex] will be produced from 237.1 g of potassium iodide
