Which volume, in cm3, of 0.20 mol dm-3 NaOH (aq) is needed to neutralize 0.050 mol of H2S(g)?

To solve this problem, we need to understand the reaction between NaOH and H2S. The balanced chemical equation is:

2NaOH + H2S -> Na2S + 2H2O

From the balanced equation, we can see that 2 moles of NaOH react with 1 mole of H2S.

Given that we have 0.050 mol of H2S, we will need 2 * 0.050 = 0.10 mol of NaOH to neutralize it.

We know that the concentration of NaOH is 0.20 mol/dm^3.

Concentration is defined as moles/volume. Therefore, we can rearrange the formula to find the volume:

Volume = moles / concentration

Substituting the given values:

Volume = 0.10 mol / 0.20 mol/dm^3 = 0.50 dm^3

Since 1 dm^3 = 1000 cm^3,

Volume = 0.50 * 1000 = 500 cm^3

Therefore, 500 cm^3 of 0.20 mol/dm^3 NaOH is needed to neutralize 0.050 mol of H2S.