Respuesta :
The given question is incomplete. The complete question is
Calculate the mass of AgCl formed, and the concentration of silver ion remaining in solution, when 10.0g of solid [tex]AgNO_3[/tex] is added to 50.mL of [tex]1.0\times 10^{-2}[/tex] NaCl. Assume there is no volume change upon addition of the solid.
Answer: a) mass of AgCl formed = 0.072 g
b) Concentration of silver ion remaining in solution
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} AgNO_3=\frac{10.0g}{169.87g/mol}=0.059moles[/tex]
[tex]\text{Moles of} NaCl=molarity\times {\text {Volume in L}}=1.0\times 10^{-2}\times 0.05L=0.0005moles[/tex]
[tex]AgNO_3(aq)+NaCl(g)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
According to stoichiometry :
1 mole of [tex]NaCl[/tex] require = 1 mole of [tex]AgNO_3[/tex]
Thus 0.0005 moles of [tex]NaCl[/tex] will require=[tex]\frac{1}{1}\times 0.0005=0.0005moles[/tex] of [tex]AgNO_3[/tex]
Thus [tex]NaCl[/tex] is the limiting reagent as it limits the formation of product and [tex]AgNO_3[/tex] is the excess reagent.
As 1 mole of NaCl give = 1 mole of AgCl
Thus 0.0005 moles of NaCl give =[tex]\frac{1}{1}\times 0.0005=0.0005moles[/tex] of [tex]H_2O[/tex]
Mass of [tex]AgCl=moles\times {\text {Molar mass}}=0.0005moles\times 143.5g/mol=0.072g[/tex]
moles of AgCl left = (0.059-0.0005) = 0.0585
Concentration of [tex][Ag]^+[/tex] left in solution =[tex]\frac{moles}{Volume}=\frac{0.0585\times 1000}{50}=1.17M[/tex]
