How many grams of barium sulfate can be produced from 20.8 g of barium chloride? \( \begin{array}{l}137 \mathrm{~g} \\ 23.3 \mathrm{~g} \\ 233 \mathrm{~g} \\ 2\end{array} \) 233 g

1. Write the balanced chemical equation: \[ \mathrm{BaCl_2} + \mathrm{Na_2SO_4} \rightarrow \mathrm{BaSO_4} + 2\,\mathrm{NaCl} \] 2. Calculate the molar masses: - For \( \mathrm{BaCl_2} \): \[ M_{\mathrm{BaCl_2}} = 137.3 + 2(35.5) \approx 208.3\ \mathrm{g/mol} \] - For \( \mathrm{BaSO_4} \): \[ M_{\mathrm{BaSO_4}} = 137.3 + 32.1 + 4(16.0) \approx 233.3\ \mathrm{g/mol} \] 3. Determine the number of moles of \( \mathrm{BaCl_2} \) in 20.8 g: \[ \text{moles of } \mathrm{BaCl_2} = \frac{20.8\ \mathrm{g}}{208.3\ \mathrm{g/mol}} \approx 0.1\ \text{mol} \] 4. Use the stoichiometry of the reaction (1:1 ratio between \( \mathrm{BaCl_2} \) and \( \mathrm{BaSO_4} \)) to find the moles of \( \mathrm{BaSO_4} \) produced: \[ \text{moles of } \mathrm{BaSO_4} = 0.1\ \text{mol} \] 5. Calculate the mass of \( \mathrm{BaSO_4} \) produced: \[ \text{mass of } \mathrm{BaSO_4} = 0.1\ \text{mol} \times 233.3\ \mathrm{g/mol} \approx 23.33\ \mathrm{g} \] Rounded to 3 significant figures, the answer is: \[ 23.3\ \mathrm{g} \]