Answer:
[Ba^2+] = 0.160 M
Explanation:
First, let's calculate the moles of each reactant with the following expression:
n = M * V
moles of K2CO3 = 0.02 x 0.200 = 0.004 moles
moles of Ba(NO3)2 = 0.03 x 0.400 = 0.012 moles
Now, let's write the equation that it's taking place. If it's neccesary, we will balance that.
Ba(NO3)2 + K2CO3 --> BaCO3 + 2KNO3
As you can see, 0.04 moles of K2CO3 will react with only 0.004 moles of Ba(NO3) because is the limiting reactant. Therefore, you'll have a remanent of
0.012 - 0.004 = 0.008 moles of Ba(NO3)2
These moles are in total volume of 50 mL (30 + 20 = 50)
So finally, the concentration of Ba in solution will be:
[Ba] = 0.008 / 0.050 = 0.160 M
