Starting with the following equation, BaCl2(aq) + Na,PO.(aq) → Ba3(PO4)2(s) + NaCI(aq) calculate the mass in grams of BaCl, that will be required to produce 69.5 grams of Ba3(PO4)2.

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### Chemical Reaction Stoichiometry Calculation

To determine the mass of \(\text{BaCl}_2\) needed to produce 69.5 grams of \(\text{Ba}_3(\text{PO}_4)_2\), start with the balanced chemical equation:

\[
\text{3BaCl}_2(\text{aq}) + \text{2Na}_3\text{PO}_4(\text{aq}) \rightarrow \text{Ba}_3(\text{PO}_4)_2(\text{s}) + \text{6NaCl}(\text{aq})
\]

### Steps for Calculation:

1. **Molar Mass Calculation:**
   - Calculate the molar mass of \(\text{Ba}_3(\text{PO}_4)_2\).
     \[
     \text{Ba}_3(\text{PO}_4)_2: 3(137.33) + 2(30.97 + 4(16)) = 601.93 \, \text{g/mol}
     \]

2. **Convert Grams to Moles:**
   - Convert grams of \(\text{Ba}_3(\text{PO}_4)_2\) to moles.
     \[
     \text{Moles of Ba}_3(\text{PO}_4)_2 = \frac{69.5 \, \text{g}}{601.93 \, \text{g/mol}} \approx 0.115 \, \text{moles}
     \]

3. **Use Stoichiometry to Find Moles of \(\text{BaCl}_2\):**
   - Use the mole ratio from the balanced equation to find moles of \(\text{BaCl}_2\).
     \[
     \frac{3 \, \text{moles BaCl}_2}{1 \, \text{mole Ba}_3(\text{PO}_4)_2} = 0.345 \, \text{moles of BaCl}_2
     \]

4. **Calculate Mass of \(\text{BaCl}_2\):**
   - Find the molar mass of \(\text{BaCl}_2\).
     \[
     \text{BaCl}_2: 137.33 + 2(35.45) = 208.23 \, \text{g/mol
Transcribed Image Text:### Chemical Reaction Stoichiometry Calculation To determine the mass of \(\text{BaCl}_2\) needed to produce 69.5 grams of \(\text{Ba}_3(\text{PO}_4)_2\), start with the balanced chemical equation: \[ \text{3BaCl}_2(\text{aq}) + \text{2Na}_3\text{PO}_4(\text{aq}) \rightarrow \text{Ba}_3(\text{PO}_4)_2(\text{s}) + \text{6NaCl}(\text{aq}) \] ### Steps for Calculation: 1. **Molar Mass Calculation:** - Calculate the molar mass of \(\text{Ba}_3(\text{PO}_4)_2\). \[ \text{Ba}_3(\text{PO}_4)_2: 3(137.33) + 2(30.97 + 4(16)) = 601.93 \, \text{g/mol} \] 2. **Convert Grams to Moles:** - Convert grams of \(\text{Ba}_3(\text{PO}_4)_2\) to moles. \[ \text{Moles of Ba}_3(\text{PO}_4)_2 = \frac{69.5 \, \text{g}}{601.93 \, \text{g/mol}} \approx 0.115 \, \text{moles} \] 3. **Use Stoichiometry to Find Moles of \(\text{BaCl}_2\):** - Use the mole ratio from the balanced equation to find moles of \(\text{BaCl}_2\). \[ \frac{3 \, \text{moles BaCl}_2}{1 \, \text{mole Ba}_3(\text{PO}_4)_2} = 0.345 \, \text{moles of BaCl}_2 \] 4. **Calculate Mass of \(\text{BaCl}_2\):** - Find the molar mass of \(\text{BaCl}_2\). \[ \text{BaCl}_2: 137.33 + 2(35.45) = 208.23 \, \text{g/mol
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