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A gaseous mixture of composition by volume, 78% H2 and 22% CO is contained in a vessel. It is desired that the mixture should be made in proportion 52% H2 and 48% CO by removing some of the mixture and adding some CO. Calculate per mole of mixture the mass of mixture to be removed, and mass of CO to be added. Assume that the pressure and temperature in the vessel remain constant during the procedure. |
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Answer» Composition of mixture by volume : H2 = 78%, CO = 22% Final composition desired : H2 = 52%, CO = 48% Since the pressure and temperature remain constant, then the number of moles in the vessel remain the same throughout. ∴ Moles of mixture removed = Moles of CO added. Let x kg of mixture be removed and y kg of CO be added. For the mixture, M = \(\sum\cfrac{V_i}VM_i\) M = 0.78 × 2 + 0.22 × 28 = 7.72 Also from equation, n = \(\cfrac mM\), we have Moles of mixture removed = \(\cfrac{x}{7.72}\) = moles of CO added = \(\cfrac y{28}\) From equation, \(\cfrac{V_i}V\) = \(\cfrac{n_i}n\), we have Moles of H2 in the mixture removed = 0.78 × \(\cfrac{x}{7.72}\) = 0.101 x Moles of H2 initially = 0.78 × 1 = 0.78 Hence, Moles of H2 remaining in vessel = 0.78 – 0.101 x But 1 mole of the new mixture is 52% H2 and 48% CO, therefore 0.78 – 0.101 x = 0.52 ∴ 0.101 x = 0.26 or x = 2.57 Mass of mixture removed = 2.57 kg Also since \(\cfrac{x}{7.72}\) = \(\cfrac y{28}\) y = \(\cfrac{28}{7.72}\) \(\times\) x = \(\cfrac{28}{7.72}\) x 2.57 = 9.32 Mass of CO added = 9.32 kg. |
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