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

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Let the precipitation reaction be

$$
\text{BaCl}_2 + \text{Na}_2\text{SO}_4 \rightarrow \text{BaSO}_4 + 2\,\text{NaCl}.
$$

Since the stoichiometric ratio between $$\text{BaCl}_2$$ and $$\text{BaSO}_4$$ is $$1:1$$, the number of moles of $$\text{BaCl}_2$$ needed equals the number of moles of $$\text{BaSO}_4$$ produced.

1. Calculate the number of moles of $$\text{BaSO}_4$$ in 100. grams. The molar mass of $$\text{BaSO}_4$$ is approximately

$$
   M(\text{BaSO}_4) = 137.3 + 32.1 + 4(16.0) \approx 233.3 \text{ g/mol}.
   $$

Thus, the moles of $$\text{BaSO}_4$$ are

$$
   n(\text{BaSO}_4) = \frac{100.0 \text{ g}}{233.3 \text{ g/mol}} \approx 0.4286\, \text{mol}.
   $$

2. Since $$1$$ mole of $$\text{BaCl}_2$$ produces $$1$$ mole of $$\text{BaSO}_4$$, you need $$0.4286$$ moles of $$\text{BaCl}_2$$.

3. Next, calculate the mass of $$\text{BaCl}_2$$ needed. The molar mass of $$\text{BaCl}_2$$ is approximately

$$
   M(\text{BaCl}_2) = 137.3 + 2(35.45) \approx 137.3 + 70.9 = 208.2 \text{ g/mol}.
   $$

Thus, the mass required is

$$
   m(\text{BaCl}_2) = 0.4286 \text{ mol} \times 208.2 \text{ g/mol} \approx 89.3 \text{ g}.
   $$

Therefore, approximately $$\boxed{89.3 \text{ g}}$$ of barium chloride are needed to make 100. grams of barium sulfate.
Approximately 89.3 grams of barium chloride are needed to make 100 grams of barium sulfate.

How many grams of barium chloride are needed to make 100. grams of barium sulfate?

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