Question #194503

Determine the mass of Lead (II) Iodide produced if 9.70g of lead (II) nitrate reacts with excess sodium iodide


1
Expert's answer
2021-05-18T07:52:43-0400

Pb(NO3)2 +2NaI --->2NaNO3 + PbI2


Moles[Pb(NO3)2]=9.70331.2=0.0293molMoles[Pb(NO_3)_2]= \frac{9.70}{331.2}=0.0293mol


Moles[NaNO3]=2×0.0293=0.0586molMoles [NaNO_3]=2×0.0293=0.0586mol


Mass[NaNO3]=149.89×0.0586=8.782gramsMass [NaNO_3]= 149.89×0.0586=8.782 grams





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