In the following reaction
2 VO43- + 3 Zn + 16 H+→ 2 V2+ + 3 Zn2+ + 8 H2O
the initial rate of disappearance of VO43- was found to be 0.56 M/s. What is the initial rate of appearance of Zn2+?
-0.56 M/s
0.37 M/s
0.56 M/s
0.84 M/s
1.12 M/s
0.84 M/s
correct it is 0.84
goodluck cuz i had to trust this answer and it was right.
To find the initial rate of appearance of Zn2+, we'll use the coefficients from the balanced chemical equation you provided:
2 VO43- + 3 Zn + 16 H+ → 2 V2+ + 3 Zn2+ + 8 H2O
The stoichiometry tells us that the ratio of the rate of disappearance of VO43- to the rate of appearance of Zn2+ is 2:3. Therefore, we can set up the following proportion:
(0.56 M/s) / (x M/s) = 2 / 3
To solve for x, the initial rate of appearance of Zn2+, we can cross multiply and solve for x:
(0.56 M/s) * (3 / 2) = x
x = 0.56 M/s * 3 / 2
x = 0.84 M/s
Therefore, the initial rate of appearance of Zn2+ is 0.84 M/s.