Na2CO3 + 2AgNO3 ==> 2NaNO3 + Ag2CO3
I like to work in millimoles. If you prefer moles just divide by 1000 to find the value in mols. Working with mmols keeps all those zeros after the decimal point out of the way; like 0.05 x 0.08 = 0.004 mols or 50 x 0.08 = 4 mmols.
millimoles Na2CO3 initially = mL x M = 50 mL x 0.0800 = 4
millimoles AgNO3 initially = 25.0 x 0.200 = 5
First determine the limiting reagent (LR). Let's go through both scenarios.
a. You have 4 mmols Na2CO3. It will require 8 mmols AgNO3 and you don't have that much so AgNO3 is the LR.
b. You have 5 mmols AgNO3. It will require 2-1/2 mmols Na2CO3, You have that much Na2CO3 (and more) so AgNO3 is the LR.
Given that AgNO3 is the LR, then 4 - 2.5 = 1.5 millimols Na2CO3 remain unreacted. So (Na2CO3) = millimoles/mL = 1.5 mmols/75 mL = ?
Note that the 75 mL comes from 50 mL of Na2CO3 + 25 mL AgNO3 = 75 mL solution. Post your work if you get stuck.
What will be the molar analytical concentration of Na CO, in the solution pro duced when 25.0 mL of 0.200 M AGNO, is mixed with 50.0 mL of 0.0800 M Na CO,?
1 answer