Question

A solution of potassium chromate (K2CrO4) reacts with a solution of lead (ll) nitrate (Pb (NO3)2). If 0.250 mol of potassium chromatic is used, what mass of lead (ll) chromate (Pb CrO4) can be produced? K2CrO4 + Pb (NO3)2 -> 2KNO3 + PbCrO4

Answer 1

K2CrO4 + Pb (NO3)2 -> 2KNO3 + PbCrO4

If 0.250 mol of potassium chromatic is used, then 0.250 mol of PbCrO4 are produced.

$Mass = moles*rfm$

Mass=0.250* 323.19= 80.8g