The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to th

Question

The fizz produced when an Alka-Seltzer tablet is dissolved in water is due to the reaction between sodium bicarbonate and citric acid, C6H8O7: 3NaHCO3 (aq) + C6H8O7 (aq) --> 3CO2 (g) + 3H2O (l) + Na3C6H5O7 (aq)

a. If 2.00 g of sodium bicarbonate and 2.00 g of citric acid are allowed to react, what is the limiting reagent?

b. How many grams of cabon dioxide can theoretically form?

c. How much of the excess reagent is lef over aftr the reaction has come to completion?

d. If you perform the experiment with an actual yield of 0.43 g, what is your percent yield for the reaction?

Explanation / Answer

a)

m = 2 g of NaHCO3

MW = 84.007

mol = 2/84.007 = 0.01041 mol of NaHCO3

m = 2 g of C6H8O7

MW = 192.124

mol = 2/192.124 = 0.0238 mol of C6H8O7

NaHCO3 is clearly the limitng reactant.

b)

CO2 will form=

3 mol of NAHCO3 form 3 mol of CO2

then 1 mol :1 mol ratio; therefore

0.01041 mol of NaHCO3 will form 0.01041 CO2

mass = mol*MW = 0.01041 *44 = 0.45804 g of CO2

c)

excess reactant:

0.01041 /3 mol of citric acid reacted = 0.00347

0.01041 - 0.00347 = 0.00694 mol of Citric acid left

then

Mass = mol*MW = 0.00694*192.124 = 1.3333 g of Citric acid is left

d)

yield = 0.43

find % yield

% = real/theoretical * 100 = 0.43 / 0.45804 * 100 = 93.87%

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