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Mass of NaHCO3 sample (g): 1.20 g
Volume of carbon dioxide produced (mL): 250 mL
8. What is the molecular mass of NaHCO3?
9. How many moles of NaHCO3 are in the sample?
10. Theoretically, how many moles should be produced from 1.2 grams of baking soda?
11. How many moles of carbon dioxide is produced in the reaction? (Use the ideal gas equation, PV=nRT.
for the calculation assuming P= 1 atm; T=298 K and R=0.0821 L-atm/mol-K)
12. What is the percentage error for the experiment?
For items 13 to 20, refer to the following laboratory experiment and results. Show your complete
solution for all the questions.
A student weighed 1.25 grams of NaHCO3 pre-weighed 50 mL beaker. In the fume hood, he slowly
added concentrated HCl to the beaker containing the NaHCO3. He added a total of 4.6 mL of
concentrated HCl. During the reaction, effervescence was observed due to carbon dioxide formation.
The water by-product was removed by heating the beaker and a white crystalline salt, NaCl, is left.
Lastly, he took the weight of the residue and the beaker after cooling to room temperature.

Respuesta :

Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol

9.

(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3

10.

Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:

NaHCO3 + H{+} = Na{+} + H2O + CO2

(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory

11.

n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2

12.

(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%

Explanation:

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