Question

+

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

Explanation:

To calculate the moles :

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

$\text{Moles of potassium iodide}=\frac{237.1g}{166g/mol}=1.428moles$

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=$\frac{1}{2}\times 1.428=0.714moles$  of $PbI_2$

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