Question

Calculate the mass of AgCl formed, and the concentration of silver ion remaining in solution, when 10.0g of solid AgNO3 is added to 50.mL of 1.0x10-2 Assume there is no volume change upon addition of the solid.

Answer 1

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
`AgNO_3` is added to 50.mL of
`1.0* 10^(-2)` 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

Step-by-step explanation:

To calculate the moles :

`\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}`

`\text{Moles of} AgNO_3=(10.0g)/(169.87g/mol)=0.059moles`

`\text{Moles of} NaCl=molarity* {\text {Volume in L}}=1.0* 10^(-2)* 0.05L=0.0005moles`

`AgNO_3(aq)+NaCl(g)\rightarrow AgCl(s)+NaNO_3(aq)`

According to stoichiometry :

1 mole of
`NaCl` require = 1 mole of
`AgNO_3`

Thus 0.0005 moles of
`NaCl` will require=
`(1)/(1)* 0.0005=0.0005moles` of
`AgNO_3`

Thus
`NaCl` is the limiting reagent as it limits the formation of product and
`AgNO_3` is the excess reagent.

As 1 mole of NaCl give = 1 mole of AgCl

Thus 0.0005 moles of NaCl give =
`(1)/(1)* 0.0005=0.0005moles` of
`H_2O`

Mass of
`AgCl=moles* {\text {Molar mass}}=0.0005moles* 143.5g/mol=0.072g`

moles of AgCl left = (0.059-0.0005) = 0.0585

Concentration of
`[Ag]^+` left in solution =
`(moles)/(Volume)=(0.0585* 1000)/(50)=1.17M`