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# Pb(NO3)2 + 2KI → PbI2 + 2KNO3

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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?

Mar 29, 2020

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$$Pb(NO_3)_2 + 2KI → PbI_2 + 2KNO_3$$

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 $$Pb (NO_3)_ 2$$. !

Mar 29, 2020