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].
