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

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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:
$$
23.3\ \mathrm{g}
$$
23.3 g

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

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