NaHCO3 + CH3COOH=>CH3COONa + H2O + CO2
8.4........20g.........x......y....4.4
So you start with 20+8.4 = 28.4
You lose 4.4 which should leave you with 24.0 grams. (In case you're interested, that's the CH3COONa(that's x = 8.2 g) + H2O(that's y=1.8 g) + whatever amount the unused acetic acid (that's 14.0 g) and
14.0 + 8.2 + 1.8 = 24.0 g.
when 8.4 g of NaHCO3 is added to a solution of CH3COOH weighing 20 g .It is observrd that 4.4 g of CO2 is releasd into atmosphere and a residue is left behind .calculate the mass of residue by applying law of conservation of mass .
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