# The fizz produced when an Alka-Seltzer® tablet is dissolved in water is due to the reaction between sodium bicarbonate (NaHCO3) and citric acid (H3C6H5O7): In a certain experiment 1.45g of sodium bicarbonate and 1.45g of citric acid are allowed to react.

How many grams of the excess reactant remain after the limiting reactant is completely consumed?

3NaHCO3+H3C6H5O7-->3CO2+3H2O+Na3C6H5O7

Step-by-step explanation much appreciated with as much detail as possible so I can understand. Thanks!

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1. Use the equation you posted.
Convert 1.45 g each reagent to mols. mols = grams/molar mass

Use the coefficients in the balanced equation to convert mols of each reactant to mols CO2. It is likely that the two numbers will not be the same; obviously one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that value is the limiting reagent.

Now convert mols of the limiting reagent to mols of the non-limiting reagent (the reagent in excess), change to grams, and subtract from 1.45 g there initially. The difference is the amount that did not react. Post your work if you get stuck.

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2. I was able to figure out that NaHCO3 was the limiting reagent. And then I got that there was .7596 grams of CO2 by converting the 1.45g of NaHCO3. I tried to use 1.45 NaHCO3 to figure it out but, I didn't get the right answer.

My work:

1.45g NaHCO3(1mol NaHCO3/83.998g NaHCO3)(1mol H3C6H5O7/3mol NaHCO3)(192.059g H3C6H5O7/1mol H3C6H5O7)

Can you tell me where I went wrong?

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3. yes.You mixed the steps I gave you.
mol NaHCO3 = 1.45/84(approx) = about 0.173.

mols CO2 from NaHCO3 = 0.173 x (3/3) = 0.173
mols CO2 from H3C = 0.00756 x (3/1) = 0.0227

The smaller number is 0.0227 mol; therefore, citric acid is the limiting reagent.

Now convert 0.00756 mols citric acid to mols NaHCO3, convert mols NaHCO3 to grams and subtract from 1.45 there initially.

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