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A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Assume no change in volume of the solution. The chemical reaction that takes place is represented by the following equation. Pb(NO3)2(s) + NaI(aq) → PbI2(s) + NaNO3(aq) How many moles of PbI2 can be produced?/Identify the limiting reactant/If the actual yield was 0.550 g of PbI2, what is the percent yield?
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A 0.450 g sample of solid lead(II) nitrate is added to 250 mL of 0.250 M sodium iodide solution. Ass...
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