Question

Question 2 of 3 > The balanced equation for the reaction of aqueous Pb(ClO2), with aqueous Nal is shown below. Pb(CIO), (aq) +2 Nal(aq) PbL_(s) + 2 NaCIO, (aq) What mass of precipitate will form if 1.50 L of excess Pb(ClO ), is mixed with 0.500 L of 0.300 M Nal? Assume 100% yield and neglect the slight solubility of Pbly. mass: 45.6

Answer 1

Number of moles of NaI = molarity * volume of solution in L

Number of moles of NaI = 0.300 * 0.500 = 0.150 mole

From the balanced equation we can say that

2 mole of NaI produces 1 mole of PbI2 so

0.150 mole of NaI will produce

= 0.150 mole of NaI *(1 mole of PbI2 / 2 mole of NaI)

= 0.0750 mole of PbI2

mass of 1 mole of PbI2 = 461.01 g so

the mass of 0.0750 mole of PbI2 = 34.6 g

Therefore, the mass of PbI2 produced would be 34.6 g