1.A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb(NO3)2 by mass (d: 1.134 g/mL). How many grams of PbI2 should form? Pb(NO3)2(aq) + 2 KI(aq) PbI2(s) + 2 KNO3(aq)

Explanation:

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In this case, the undergoing chemical reaction is:

Thus, we proceed to compute the reacting moles of Pb(NO3)2 and KI, by using the given concentrations and densities and molar masses which are 331.2 g/mol and 166 g/mol respectively:

Next, the 0.0464 moles of Pb(NO3)2 will consume the following moles of KI (consider their 1:2 molar ratio):

Hence, as only 0.0789 moles of KI are available, KI is the limiting reactant, therefore the formed grams of PbI2, considering its molar mass of 461.01 g/mol and 2:1 molar ratio, are:

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