How many grams of barium chloride are needed to make 100. grams of barium sulfate? \( \begin{array}{l}46.6 \mathrm{~g} \\ 89.3 \mathrm{~g} \\ 44.9 \mathrm{~g} \\ 233.3 \mathrm{~g} \\ 208.3 \mathrm{~g}\end{array} \)

To find out how many grams of barium chloride (BaCl₂) are needed to produce 100 grams of barium sulfate (BaSO₄), we first need to look at the balanced chemical reaction: \[ \text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\text{NaCl} \] From the reaction, we see that 1 mole of barium chloride produces 1 mole of barium sulfate. The molar masses are approximately: - BaCl₂: 137.33 (Ba) + 2 × 35.45 (Cl) = 207.23 g/mol - BaSO₄: 137.33 (Ba) + 32.07 (S) + 4 × 16.00 (O) = 233.39 g/mol Now, we can find out how many moles of barium sulfate correspond to 100 grams: \[ \text{Moles of BaSO₄} = \frac{100 \text{ g}}{233.39 \text{ g/mol}} \approx 0.428 \text{ moles} \] Using the stoichiometry from the balanced equation (1:1 ratio), we also need 0.428 moles of BaCl₂ to produce this amount of BaSO₄. Now calculate the mass of BaCl₂: \[ \text{Mass of BaCl₂} = 0.428 \text{ moles} \times 207.23 \text{ g/mol} \approx 88.75 \text{ g} \] Rounding this, we can say that approximately 89.3 grams of barium chloride are needed to produce 100 grams of barium sulfate. So the answer is: 89.3 g