Answered

Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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 :

GinnyAnswer
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].
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.