If 12.8 g lead(II) sulfate (303.3 g/mol) precipitates when excess potassium chloride is added to 1.65 L of a water sample, what is molar concentration of Pb2+ in the sample?

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

In such a way, since all the lead (II) is converted due to the excess of potassium chloride, the moles of lead (II) in the sample are computed from the mass of lead (II) sulfate:

Thus, since volume of the solution is 1.65 due to the fact that the addition of the reactants is not enough to significantly modify the reaction's volume, the resulting molar concentration of the lead (II) ions is:

Regards.

> b takes place in the gas phase in an isothermal tubular reactor. the feed is one mole of a per one mole of c, an inert. the entering temperature and pressure are 427°c and 10 atm, respectively. the gas constant r = 0.08206 atm · l/mol · k.