# Question 96f8c

Mar 22, 2015

Since you've got the first part, I'll just answer the second question, so to speak.

So, you've got your balanced chemical equation for this double replacement reaction

$A g N {O}_{3 \left(a q\right)} + K C {l}_{\left(a q\right)} \to A g C {l}_{\left(s\right)} + K N {O}_{3 \left(a q\right)}$

You know that you have a $\text{1:1}$ mole ratio between all the species involved. This means that the number of moles of potassium chloride must be equal to the number of moles of silver nitrate.

Since I assume you've calculated the number of moles of silver nitrate to be

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{A g N {O}_{3}} = \text{0.162 M" * "1.27 L" = "0.206 moles}$ $A g N {O}_{3}$

automatically you'll have

$\text{0.206"cancel("moles "AgNO_3) * ("1 mole" KCl)/(cancel("1 mole "AgNO_3)) = "0.206 moles}$ $K C l$

Now use the volume of potassium chloride given to figure out what the solution's concentration must be in order to have that many moles of $K C l$ available for the reaction

C = n/V = "0.206 moles KCl"/"3.78 L" = color(red)("0.0545 M")#