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Lady_Fox [76]
3 years ago
6

Which is the balanced chemical equation representing the reaction between

Chemistry
2 answers:
Travka [436]3 years ago
5 0

Answer:

Choice A: \rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to 1\;NaC_2H_3O_2 + 1\;H_2O + 1\;CO_2.

Explanation:

Indeed it is possible to balance this equation by the conservation of atoms in a chemical reaction. However, knowing what's actually going on in this process will likely make this problem easier to solve.

Vinegar contains acetic acid \rm C_2H_4 O_2. Acetic acid is a monoprotic acid. In other words, each \rm C_2H_4 O_2 can dissociate to produce up to one hydrogen ion \rm H^{+}. That is:

\rm C_2H_4 O_2 \rightleftharpoons {C_2H_3 O_2}^{-} + H^{+}.

Baking soda is a common name for sodium bicarbonate \rm NaHCO_3. Each formula unit of \rm NaHCO_3 contains one bicarbonate ion: \rm {HCO_3}^{-}. Each bicarbonate ion will consume one hydrogen ion to produce water and carbon dioxide:

\rm {HCO_3}^{-} + H^{+} \to H_2O + CO_2.

For this chemical equation to balance, the number of hydrogen ions that \rm C_2H_4 O_2 supplies shall be the same as the number of these ions that \rm NaHCO_3 consumes. Each unit of \rm C_2H_4 O_2 supplies one unit of hydrogen ions while each unit of \rm NaHCO_3 consumes one unit of hydrogen ions. Reacting the two at a one-to-one ratio will make sure that this reaction neither run short of hydrogen ions or produce more hydrogen ions than it need.

Hence the coefficient in front of \rm C_2H_4 O_2 and \rm NaHCO_3 shall be the same. Let their coefficients be one.

\rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to ?\;NaC_2H_3O_2 + ?\;H_2O + ?\;CO_2.

Now, balance this equation with reference to the number of atoms:

  • One Na atom;
  • Five H atoms;
  • Five O atoms;
  • Three C atoms.

\rm 1\; C_2H_4 O_2 + 1\;NaHCO_3 \to 1\;NaC_2H_3O_2 + 1\;H_2O + 1\;CO_2.

Citrus2011 [14]3 years ago
4 0

Answer:

The correct option is OA.

C2H4O2 + NaHCO3 - NaC2H302 + H2O + CO2

Explanation:

To solve this you have to check the number of elements in both sides of the equation.

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