chem...equilibrium

Why 15.3 ?
No. of moles of CH3CH2OH = 0.33 mol
Molar mass of CH3CH2OH = 46 g/mol
Hence, mass of CH3CH2OH = 0.33 x 46 = 15.2 g
It is not 15.3 g.
If fact, you have to copy the no. of moles of original equilibrium (i.e. 0.33 mol).
It is not necessary to write in the form (mass/molar mass).

Why 0.77 but not 0.87 ?
Further no. of moles of CH­3CH2OH reacted = 6/60 = 0.1 mol
Mole ratio (CH3CH2OH reacted) : (CH3COOCH2CH3 formed) = 1 : 1
Further no. of moles of CH3COOCH2CH3 formed = 0.1 mol
No. of moles of CH3COOCH2CH3 in original equilibrium = 0.67 mol
No. of moles of CH3COOCH2CH3­­ in new equilibrium = 0.67 + 0.1 = 0.77 mol
Similarly, no. of moles of H2O in new equilibrium = 0.67 + 0.1 = 0.77 mol
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