The volume of hydrogen sulphide it stp when 1.5g of sodium hydroxide react with hydrogen sulphide.
(NA=23,H=1,S=32,O=16)
Solution:
The molar mass of sodium hydroxide (NaOH) is (23 + 16 + 1) = 40 g/mol
Therefore,
Moles of NaOH = (1.5 g NaOH) × (1 mol NaOH / 40 g NaOH) = 0.0375 mol NaOH
Balanced chemical equation:
2NaOH + H2S → Na2S + H2O
According to stoichiometry:
2 mol of NaOH react with 1 mol of H2S
Thus 0.0375 mol of NaOH react with:
(0.0375 mol NaOH) × (1 mol H2S / 2 mol NaOH) = 0.001875 mol H2S
At STP, one mole of any gas occupies a volume of 22.4 L
Thus 0.001875 mol of H2S occupy:
(0.001875 mol H2S) × (22.4 L H2S / 1 mol H2S) = 0.42 L H2S
Answer: The volume of hydrogen sulphide (H2S) is 0.42 L
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