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We wish to determine the mass of [tex]BaSO_4[/tex] formed when [tex]200 \, \text{mL}[/tex] of [tex]0.10 \, M \, BaCl_2[/tex] reacts with excess [tex]Na_2SO_4[/tex] according to the equation below.

[tex]\[
BaCl_2(aq) + Na_2SO_4(aq) \rightarrow BaSO_4(s) + 2NaCl(aq)
\][/tex]

How many moles of [tex]BaCl_2[/tex] are present in [tex]200 \, \text{mL}[/tex] of [tex]0.10 \, M \, BaCl_2[/tex]?

Moles of [tex]BaCl_2[/tex]:

Sagot :

Let's walk through the process of determining the moles of [tex]\( \text{BaCl}_2 \)[/tex] present in 200 mL of 0.10 M [tex]\( \text{BaCl}_2 \)[/tex].

1. Volume Conversion:
- First, convert the volume from milliliters (mL) to liters (L) since molarity is given in terms of moles per liter (M).
- 200 mL [tex]\( \times \frac{1 \text{ L}}{1000 \text{ mL}} = 0.2 \text{ L} \)[/tex]

2. Calculate the Moles of [tex]\( \text{BaCl}_2 \)[/tex]:
- Use the molarity (0.10 M) to find the moles of [tex]\( \text{BaCl}_2 \)[/tex] in the 0.2 liters of solution.
- Molarity (M) is defined as moles of solute per liter of solution.
- Therefore, moles of [tex]\( \text{BaCl}_2 \)[/tex] = Molarity [tex]\( \times \)[/tex] Volume (in liters)
- [tex]\( \text{moles of BaCl}_2 = 0.10 \text{ M} \times 0.2 \text{ L} = 0.020 \text{ mol} \)[/tex]

Thus, there are [tex]\( 0.020 \)[/tex] moles of [tex]\( \text{BaCl}_2 \)[/tex] present in 200 mL of 0.10 M [tex]\( \text{BaCl}_2 \)[/tex].