A 2.34 g sample of pure NaCl is heated with excess solid $K X_{2} Cr X_{2} O X_{7}$ and conc. $H X_{2} SO X_{4}$ to give $CrO X_{2} Cl X_{2}$ ( $4 NaCl + K X_{2} Cr X_{2} O X_{7} + 6 H X_{2} SO X_{4} 2 CrO X_{2} Cl X_{2} + 4 NaHSO X_{4} + 2 KHSO X_{4} + 3 H X_{2} O$ ). The vapours are absorbed in NaOH to give sodium chromate, and excess lead acetate is added to give $PbCrO X_{4}$ . Taking an overall yield of 80 percent from NaCl to $PbCrO X_{4}$ , calculate the mass of $PbCrO X_{4}$ formed in grams, correct to two decimal places. Atomic masses: Na 23, Cl 35.5, Cr 52, Pb 207, O 16.

Numerical answer type, answer in g, tolerance 0.01.

Answer: 5.17 g

Moles NaCl = 2.34/58.5 = 0.0400. From the equation, 4 mol NaCl give 2 mol

CrO X_{2} Cl X_{2}

, so 0.0200 mol

CrO X_{2} Cl X_{2}

, hence 0.0200 mol

Na X_{2} CrO X_{4}

and

PbCrO X_{4}

(M = 207 + 52 + 64 = 323) at 100 percent yield. Mass at 80 percent yield = 0.0200 x 323 x 0.80 = 5.168, i.e. 5.17 g.