Question

For the reaction 2 Kl+Pb(NO 3 ) 2 Pbl 2 +2 KNO 3 how many grams of lead(II) nitrate, Pb(NO 3 ) 2 are needed to react completely with 71.9 g of potassium iodide, KI?

Answer 1

71.73grams

Explanations:

Given the chemical reaction;

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

Given the following parameter

Mass of KI = 71.9grams

Determine the moles of KI

According to stoichiometry, 2moles of KI reacts with 1 mole of lead(II) nitrate,the moles of lead(II) nitrate that reacted is given as:

`\begin{gathered} moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=(1)/(2)*0.4331 \\ moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166moles \end{gathered}`

Determine the mass of lead(II) nitrate

`\begin{gathered} mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=moles* molar\text{ mass} \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166*331.2 \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=71.73g \end{gathered}`

Hence the mass of lead(II)nitrate needed to react completely with 71.9 g of potassium iodide is 71.73grams