Asked by Faith

A particular vinegar contains 4.0% CH3COOH by mass. It reacts with sodium carbonate to produce sodium acetate, carbon dioxide, and water. How many grams of carbon dioxide are produced by the reaction of 5.00 mL of this vinegar with an excess of sodium carbonate? The density of the vinegar is 1.01g/mL.
Work so far:
5.00mL vinegar x 1.01g vinegar/1mL vinegar = 5.05g vinegar
5.05g CH3COOH x 4.0g vinegar/100g vinegar = 0.202g CH3COOH
After that I have no clue. How am I supposed to figure stoichiometric coefficients and mole ratios if I don't know what is in the other 96% of the vinegar? And how do I balance the equation without the vinegar's formula? Or does ONLY the CH3COOH react with sodium carbonate? I am confused. Please help!!
Answered by DrBob222
Most if not all of the "other" material is water and the assumption is that only the vinegar reacts with the Na2CO3.
Na2CO3 + 2HOOCCH3 ==> H2O + CO2 + 2NaOOCCH3

What you have so far looks good. The above gives you the balanced equation. That little bump may be all that was holding you from finishing.
Answered by Faith
Thank you so much!
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