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?
+14913 \(Pb(NO_3)_2 + 2KI → PbI_2 + 2KNO_3 \)
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?
Hello Guest!
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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