Pb(NO3)2 + 2KI → PbI2 + 2KNO3

$Pb(NO_3)_2 + 2KI → PbI_2 + 2KNO_3$

For the reaction represented by the equation Pb(NO3)2 + 2KI → PbI2 + 2KNO3, how many moles of lead iodide are produced from 300. g of potassium iodide when Pb(NO3)2 is in excess?

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relative atomic mass:

K  → 39.0983u

I   → 126.90447u

Pb→ 206u

$2KI → PbI_2\\ 2\cdot(39.0983u+126.90447u)→ 206u+2\cdot 126.90447u\\ \color{blue}332.00554u→459.80894u$

$300g : 332.00554 = x :459.80894u\\ x=\frac{300g\cdot 459.80894u}{332.00554u}$

x = 415.483g

$1u=1.661\cdot 10^{-24}g\\ KI/1mol=166.00277u\cdot 1.661\cdot 10^{-24}g/mol=2.7573\cdot 10^{-24}g/mol$

$415.483g/(2.7573\cdot 10^{-24}g/mol)=\color{blue}1.50684399388\cdot 10^{24}mol$

415,483g KI ($1.50684399388\cdot 10^{24}mol\ KI$ )

are produced from 300g KI and sufficient $Pb (NO_3)_ 2$.

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