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:

Hello,

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:

Best regards.

if you're asking what comes out, it's steam.

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.