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The word equation for this reaction can be stated as:
sodium bicarbonate + citric acid -> sodium citrate + carbon dioxide +water
We can also write this into balanced chemical equation:
`3NaHCO_3_(aq)+H_3C_6H_5O_7_(aq) -> 3H_2O(l) + Na_3C_6H_5O_7_(aq) + 3CO_2_(g)`
Since we only know the amount of sodium bicarbonate in the problem, we assume that citric acid is in excess.
To solve for the mass of water produced, we will start with the amount of sodium bicarbonate.
`0.42 grams NaHCO_3 * (1 mol e NaHCO_3)/(84.007 grams) * (3 mol es H_2O)/(3 mol es NaHCO_3) * (18.0 grams H_2O)/(1 mol e H_2O)`
= 0.08999 = 0.090 grams H2O is produced.
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