Please check your answer. My dear friend, I hope you will got your answer to your question.:
Explanation:
The stoichiometry of the reaction tells us that 3 moles of B are consumed to produce 2 moles of C. Therefore, the rate at which C appears is directly related to the rate at which B disappears.
If B disappears at 0.6 M/s, then the rate of appearance of C can be calculated using the stoichiometric ratio.
Given that the stoichiometric coefficient of B is 3 and that of C is 2, we can say:
\[\text{Rate of } C = \frac{2}{3} \times \text{Rate of } B\]
\[\text{Rate of } C = \frac{2}{3} \times 0.6 \, \text{M/s}\]
\[\text{Rate of } C = 0.4 \, \text{M/s}\]
So, C appears at a rate of 0.4 M/s.
