College

Starting with the following equation:

[tex]\[ \text{BaCl}_2(aq) + \text{Na}_3\text{PO}_4(aq) \rightarrow \text{Ba}_3\left(\text{PO}_4\right)_2(s) + \text{NaCl}(aq) \][/tex]

Calculate the mass in grams of [tex]\[\text{BaCl}_2\][/tex] that will be required to produce 46.5 grams of [tex]\[\text{Ba}_3\left(\text{PO}_4\right)_2\][/tex].

Answer :

To solve this problem, we'll determine how much barium chloride ([tex]\(BaCl_2\)[/tex]) is needed to produce 46.5 grams of barium phosphate ([tex]\(Ba_3(PO_4)_2\)[/tex]) according to the given chemical equation:

[tex]\[ BaCl_2(aq) + Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + NaCl(aq) \][/tex]

### Step-by-Step Solution:

1. Write the Balanced Chemical Equation:

First, make sure the chemical equation is balanced. The balanced equation is:

[tex]\[ 3BaCl_2(aq) + 2Na_3PO_4(aq) \rightarrow Ba_3(PO_4)_2(s) + 6NaCl(aq) \][/tex]

From this equation, we see that 3 moles of [tex]\(BaCl_2\)[/tex] produce 1 mole of [tex]\(Ba_3(PO_4)_2\)[/tex].

2. Calculate the Molar Masses:

- Molar mass of [tex]\(Ba_3(PO_4)_2\)[/tex]:
- Ba: 137.33 g/mol
- P: 30.97 g/mol
- O: 15.999 g/mol

[tex]\[
\text{Molar mass of } Ba_3(PO_4)_2 = 3 \times 137.33 + 2 \times (30.97 + 4 \times 15.999)
\][/tex]

- Molar mass of [tex]\(BaCl_2\)[/tex]:
- Ba: 137.33 g/mol
- Cl: 35.45 g/mol

[tex]\[
\text{Molar mass of } BaCl_2 = 137.33 + 2 \times 35.45
\][/tex]

3. Calculate the Moles of [tex]\(Ba_3(PO_4)_2\)[/tex]:

Given that 46.5 grams of [tex]\(Ba_3(PO_4)_2\)[/tex] is produced, we use its molar mass to find the number of moles:

[tex]\[
\text{Moles of } Ba_3(PO_4)_2 = \frac{46.5 \text{ grams}}{\text{Molar mass of } Ba_3(PO_4)_2}
\][/tex]

4. Determine the Moles of [tex]\(BaCl_2\)[/tex] Needed:

According to the balanced equation, 3 moles of [tex]\(BaCl_2\)[/tex] are required to produce 1 mole of [tex]\(Ba_3(PO_4)_2\)[/tex]. Therefore, multiply the moles of [tex]\(Ba_3(PO_4)_2\)[/tex] by 3:

[tex]\[
\text{Moles of } BaCl_2 \text{ needed} = \text{Moles of } Ba_3(PO_4)_2 \times 3
\][/tex]

5. Calculate the Mass of [tex]\(BaCl_2\)[/tex] Needed:

Finally, convert the moles of [tex]\(BaCl_2\)[/tex] back to grams using the molar mass of [tex]\(BaCl_2\)[/tex]:

[tex]\[
\text{Mass of } BaCl_2 = \text{Moles of } BaCl_2 \times \text{Molar mass of } BaCl_2
\][/tex]

Using these steps, we find that you will need approximately 48.26 grams of [tex]\(BaCl_2\)[/tex] to produce 46.5 grams of [tex]\(Ba_3(PO_4)_2\)[/tex].