Astudent-made sample of sodium carbonate weighing 0.169g that generates 37.6 ml of co2 gas when treated with sulfuric acid. the atmospheric pressure in the lab is 731.8 mm hg and the temp. is 21.3-degree celsius. a. what is the partial pressure of co2 gas generated from the sample in mm hg? in atm? b. how many moles of co2 gas are produced? c. how many moles of pure na2co3 were there in the sample? d. how much was pure na2co3 there in the sample, in grams? e. what was the percent of pure na2co3 in the sample?

First let us consider the CO₂ formed in the reaction where the variables are:

>Pressure, P = (731.8/760) atm

>Volume, V = 37.6 mL = 0.0376 L

>Gas constant, R = 0.0821 atm L / (mol K)

>Temperature, T = (273.2 + 21.3) K = 294.5 K

From the ideal gas law : PV = nRT

Then, n = PV/(RT)

number of moles CO₂ formed, n = (731.8/760) × 0.0376 / (0.0821 × 294.5) mol = 0.001497 mol

The complete balanced chemical equation for the reaction is:

Na₂CO₃ + H₂SO₄ → Na₂SO₄ + H₂O + CO₂

We see that the mole ratio is Na₂CO₃ : CO₂ = 1 : 1

Hence, the number of moles of Na₂CO₃ which reacted

 => n = 0.001497 mol

Calculating for molar mass of Na₂CO₃

 => (23.0×2 + 12.0 + 16.0×2) g/mol = 106.0 g/mol

Therefore, mass of Na₂CO₃ reacted

=> (0.001497 mol) × (106.0 g/mol) = 0.159 g

So the percent purity of Na₂CO₃ in the sample is

=> (0.159/0.169) × 100% = 94.1%

Summary of answers:

a. 731.8 mm Hg; 0.963 atm

b. 0.001497 mol

c. 0.001497 mol

d. 0.159 g

e. 94.1%

explanation: