It is not used up in a reaction. This is because a catalyst is added to a reaction in order to speed it up, but only for that—it doesn’t react with anything, so it isn’t a product nor reactant of any reaction
Answer:
14.2 L H2S
Explanation:
Mole ratio
2 mol H2S:2 mol SO2
We are just looking for volume, so the required amount is the same as the produced amount as long as the mole ratio is balanced.
N (H2O) = m/M
m (H2O) = 1.4 kg = 1400g
M (H2O) = 2 x 1 + 16 = 18g/mol
n (H2O) = 1400/18 = 77.8mol
NaOH is a strong base and complete dissociation into Na⁺ and OH⁻ ions.
Therefore [NaOH] = [OH⁻]
To calculate the [OH⁻], we can first find the pOH as NaOH is a basic solution.
pH + pOH = 14
Since pH = 11.50
pOH = 14 - 11.50
pOH = 2.50
We can calculate [OH⁻] by knowing pOH
pOH = -log[OH⁻]
[OH⁻] = antilog(-pOH)
[OH⁻] = 3.2 x 10⁻³ M
therefore [NaOH] = 3.2 x 10⁻³ M