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!!
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.
Thank you so much!
To find the number of grams of carbon dioxide produced by the reaction, we need to first balance the equation and then determine the stoichiometric coefficients and mole ratios. You are correct that we need to know the formula of vinegar to balance the equation and determine the stoichiometry. However, we can assume that only acetic acid (CH3COOH) reacts with sodium carbonate, since it is the main component responsible for the acidity of vinegar.
The balanced equation for the reaction between acetic acid and sodium carbonate is:
2 CH3COOH + Na2CO3 -> 2 CH3COONa + H2O + CO2
From the balanced equation, we can see that 2 moles of acetic acid reacts with 1 mole of sodium carbonate to produce 1 mole of carbon dioxide.
Now let's calculate the number of moles of acetic acid in 5.05g:
Molar mass of CH3COOH = 2(12.01g) + 2(1.01g) + 16.00g = 60.05g/mol
Number of moles of CH3COOH = mass / molar mass = 0.202g / 60.05g/mol = 0.00336 mol
Since 2 moles of acetic acid reacts with 1 mole of carbon dioxide, we can determine the number of moles of carbon dioxide produced:
Number of moles of CO2 = 0.00336 mol / 2 = 0.00168 mol
To find the mass of carbon dioxide produced, we need to multiply the number of moles by the molar mass of CO2:
Molar mass of CO2 = 12.01g + 2(16.00g) = 44.01g/mol
Mass of CO2 = number of moles x molar mass = 0.00168 mol x 44.01g/mol = 0.074g
Therefore, the reaction of 5.00 mL of vinegar with an excess of sodium carbonate produces approximately 0.074 grams of carbon dioxide.
To solve this problem, you will need to use the balanced chemical equation for the reaction between CH3COOH (acetic acid) and sodium carbonate (Na2CO3). From the equation, you can determine the stoichiometric coefficients and mole ratios.
Let's start by writing the balanced equation for the reaction:
2 CH3COOH + Na2CO3 -> 2 CH3COONa + CO2 + H2O
From the balanced equation, you can see that 2 moles of acetic acid (CH3COOH) react with 1 mole of sodium carbonate (Na2CO3) to produce 1 mole of carbon dioxide (CO2).
Now let's calculate the number of moles of acetic acid (CH3COOH) in 5.05 grams. To do this, you need to divide the mass of acetic acid by its molar mass.
The molar mass of CH3COOH is calculated as follows:
(1 carbon atom x 12.01 g/mol) + (2 hydrogen atoms x 1.01 g/mol) + (2 oxygen atoms x 16.00 g/mol) + (1 oxygen atom x 12.01 g/mol) = 60.05 g/mol
Now you can calculate the number of moles of acetic acid:
moles of CH3COOH = mass of CH3COOH / molar mass of CH3COOH
moles of CH3COOH = 0.202g / 60.05 g/mol = 0.00337 mol
Since the balanced equation tells us that 2 moles of acetic acid (CH3COOH) react to produce 1 mole of carbon dioxide (CO2), we can use the mole ratio to calculate the moles of carbon dioxide produced.
moles of CO2 = 0.00337 mol of CH3COOH x (1 mol of CO2 / 2 mol of CH3COOH) = 0.00169 mol
Finally, we can convert the moles of carbon dioxide to grams.
grams of CO2 = moles of CO2 x molar mass of CO2
grams of CO2 = 0.00169 mol x (1 carbon atom x 12.01 g/mol + 2 oxygen atoms x 16.00 g/mol) = 0.0803 g
Therefore, the reaction of 5.00 mL of this vinegar with an excess of sodium carbonate will produce approximately 0.0803 grams of carbon dioxide.