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