Join IDNLearn.com and start getting the answers you've been searching for. Our Q&A platform offers detailed and trustworthy answers to ensure you have the information you need.

What volume of [tex]0.100 \, M \, AgNO_3[/tex] solution is required to completely react with the [tex]CaCl_2[/tex] in 3.00 L of a [tex]0.100 \, M \, CaCl_2[/tex] solution? The balanced chemical equation is:

[tex]\[ 2 AgNO_3 + CaCl_2 \rightarrow 2 AgCl + Ca(NO_3)_2 \][/tex]

[tex]\[\square \, L \, AgNO_3\][/tex]


Sagot :

To determine the volume of 0.100 M AgNO[tex]\(_3\)[/tex] solution needed to completely react with the CaCl[tex]\(_2\)[/tex] in 3.00 liters of a 0.100 M CaCl[tex]\(_2\)[/tex] solution, follow these steps:

1. Calculate the moles of CaCl[tex]\(_2\)[/tex] in the given solution:

We know that concentration (Molarity [tex]\(M\)[/tex]) is defined as moles of solute per liter of solution. From the problem, the concentration of CaCl[tex]\(_2\)[/tex] is 0.100 M and the volume is 3.00 liters.

[tex]\[ \text{Moles of CaCl}_2 = \text{concentration} \times \text{volume} = 0.100 \, \text{M} \times 3.00 \, \text{L} = 0.300 \, \text{moles} \][/tex]

2. Use the balanced chemical equation to find the moles of AgNO[tex]\(_3\)[/tex] required:

The balanced chemical equation is:
[tex]\[ 2 \, \text{AgNO}_3 + \text{CaCl}_2 \rightarrow 2 \, \text{AgCl} + \text{Ca(NO}_3)_2 \][/tex]

This equation tells us that 1 mole of CaCl[tex]\(_2\)[/tex] reacts with 2 moles of AgNO[tex]\(_3\)[/tex]. Therefore, to react completely with 0.300 moles of CaCl[tex]\(_2\)[/tex]:

[tex]\[ \text{Moles of AgNO}_3 \, \text{needed} = 2 \times \text{moles of CaCl}_2 = 2 \times 0.300 \, \text{moles} = 0.600 \, \text{moles} \][/tex]

3. Calculate the volume of 0.100 M AgNO[tex]\(_3\)[/tex] solution required:

Again, using the definition of molarity, we can rearrange the formula to find the volume:

[tex]\[ \text{Volume of AgNO}_3 \, \text{solution} = \frac{\text{moles of AgNO}_3}{\text{concentration of AgNO}_3} = \frac{0.600 \, \text{moles}}{0.100 \, \text{M}} = 6.00 \, \text{liters} \][/tex]

Therefore, the volume of 0.100 M AgNO[tex]\(_3\)[/tex] solution required to completely react with the CaCl[tex]\(_2\)[/tex] in 3.00 liters of 0.100 M CaCl[tex]\(_2\)[/tex] solution is [tex]\( \boxed{6.00 \, \text{liters}} \)[/tex].
Thank you for using this platform to share and learn. Keep asking and answering. We appreciate every contribution you make. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.