Respuesta :
Answer:
71.73grams
Explanations:
Given the chemical reaction;
[tex]2KI+Pb(NO_3)_2\rightarrow PbI_2+2KNO_3[/tex]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:
[tex]\begin{gathered} moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=\frac{1}{2}\times0.4331 \\ moles\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166moles \end{gathered}[/tex]Determine the mass of lead(II) nitrate
[tex]\begin{gathered} mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=moles\times molar\text{ mass} \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=0.2166\times331.2 \\ mass\text{ of Pb\lparen NO}_3\text{\rparen}_2=71.73g \end{gathered}[/tex]Hence the mass of lead(II)nitrate needed to react completely with 71.9 g of potassium iodide is 71.73grams
