18 g sulfur-containing compound produces 1.02 g BaSO4. What is the percentage of sulfur in the compound?

Step 1: Molar mass of $\ce{BaSO4}$
$M(\ce{BaSO4}) = 137 + 32 + 64 = 233~\mathrm{g/mol}$

Step 2: Mass of sulfur in $1.02~\mathrm{g}$ of $\ce{BaSO4}$
Formula:
$\mathrm{mass~of~S} = \frac{32}{233} \times \mathrm{mass~of~BaSO4}$
Substitute:
$\mathrm{mass~of~S} = \frac{32}{233} \times 1.02~\mathrm{g}$
Calculate $\frac{32}{233}$:
$\frac{32}{233} \approx 0.13734$
Then:
$0.13734 \times 1.02 = 0.1401~\mathrm{g}$

Step 3: Percentage of sulfur in the compound
Formula:
$\mathrm{\%S} = \left( \frac{mass~of~S}{mass~of~compound} \right) \times 100$
Substitute:
$\mathrm{\%S} = \left( \frac{0.1401~g}{2.18~g} \right) \times 100$
Simplify:
$\frac{0.1401}{2.18} \approx 0.06427$
$0.06427 \times 100 = 6.43\% \approx 6.42\%$

Answer: (d) $6.42\%$
6.42%