First, you need to calculate the standard cell potential using standard reduction potential from a textbook or online. Since Mg becomes Mg+2, magnesium is being oxidized because it is losing electrons, you need to flip its potential
Fe+2 + 2e- --> Fe potential= -0.44
Mg+2 + 2e- --> Mg potential= -2.37
Cell potential= (-0.44) + (+2.37)= 1.93 V
Now, you need to use Nernst formula to get the answer. I have attached a PDF with the work.
Answer:
Qsp > Ksp, BaCO3 will precipitate
Explanation:
The equation of the reaction is;
Na2CO3 + BaBr2 -------> 2NaBr + BaCO3
Since BaCO3 may form a precipitate we can determine the Qsp of the system.
Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles
concentration of NaCO3 = number of moles/volume of solution = 9.1 * 10^-3 moles/10 L = 9.1 * 10^-4 M
Number of moles of BaBr2 = 0.20g/297 g/mol = 6.7 * 10^-4 moles
concentration of BaBr2 = 6.7 * 10^-4 moles/10 L = 6.7 * 10^-5 M
Hence;
[Ba^2+] = 6.7 * 10^-5 M
[CO3^2-] = 9.1 * 10^-4 M
Qsp = [6.7 * 10^-5] [9.1 * 10^-4]
Qsp = 6.1 * 10^-8
But, Ksp for BaCO3 is 5.1*10^-9.
Since Qsp > Ksp, BaCO3 will precipitate
First, we have to calculate the number of moles of H2SO4 in the solution:
V=60 mL = 0.06 L
c=5.85 mol/L
n=V×c=0.06×5.85=0.351 mol
Then we need to find the molar mass of H2SO4:
2×Ar(H) + Ar(S) + 4×Ar(O) =
=2 + 32 + 64 = 98 g/mol
Finally, we need to find the mass of H2SO4:
m=0.351 × 98 = 34.398 g
