In a chromyl chloride test, 1.30 g of $NaCl$ is heated with excess $K X_{2} Cr X_{2} O X_{7}$ and conc. $H X_{2} SO X_{4}$ : $K X_{2} Cr X_{2} O X_{7} + 4 NaCl + 6 H X_{2} SO X_{4} 2 CrO X_{2} Cl X_{2} + 2 KHSO X_{4} + 4 NaHSO X_{4} + 3 H X_{2} O$ . All the $CrO X_{2} Cl X_{2}$ is absorbed in $NaOH$ to give chromate, which is precipitated completely with excess lead acetate as $PbCrO X_{4}$ . Assuming 100 percent yield in every step, find the mass of $PbCrO X_{4}$ in grams to two decimal places. (Na 23, Cl 35.5, Pb 207, Cr 52, O 16)

Numerical answer type, answer in g, tolerance 0.01.

Answer: 3.59 g

4 mol

NaCl

give 2 mol

CrO X_{2} Cl X_{2}

, and each

CrO X_{2} Cl X_{2}

gives 1 mol

CrO X_{4} X^{2 -}

and then 1 mol

PbCrO X_{4}

. So 2 mol

NaCl

give 1 mol

PbCrO X_{4}

n (NaCl) = 1.30/58.5 = 0.02222

mol, giving 0.01111 mol

PbCrO X_{4}

(molar mass 323). Mass

= 3.59