Question

Mass of NaHCO3 sample (g): 1.20 g

Answer 1

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: