Question
When one half teaspoon of baking soda is dissolved in your stomach, what volume of carbon dioxide gas would be produced at body temperature and 1.00 atm. pressure? On the average, one half teaspoon of baking soda contains 2.40 g of sodium bicarbonate. Assume the net ionic reaction, H+ + HCO3- ==> H2O(l) + CO2(g), goes to completion at these stomach conditions.
Answers
DrBob222
We assume there is enough acid in the stomach to react with all of the NaHCO3; i.e., NaHCO3 is the limiting reagent.
mols NaHCO3 = grams/molar mass = ?
1 mol NaHCO3 produces 1 mol CO2 according to your equation; therefore, mols NaHCO3 - mols CO2 produced.
Then use PV = nRT. YOu know P, solve for V, n = from above, R and T you know. Remember to use T in kelvin. V will be in liters.
mols NaHCO3 = grams/molar mass = ?
1 mol NaHCO3 produces 1 mol CO2 according to your equation; therefore, mols NaHCO3 - mols CO2 produced.
Then use PV = nRT. YOu know P, solve for V, n = from above, R and T you know. Remember to use T in kelvin. V will be in liters.
sea lively
So will the mol value for PV=nRT be 0?
DrBob222
No, and your gut feeling should tell you that is not right for a zero for n makes V = 0 and you know that can't be right
Look at my post. You calculate mols NaHCO3 from the mass at the begining and mols CO2 then = mols NaHCO3. That is n.
Look at my post. You calculate mols NaHCO3 from the mass at the begining and mols CO2 then = mols NaHCO3. That is n.
sea lively
Because mol NaHCO3=mol CO2 produced
sea lively
If not, how can I find mol CO2
sea lively
Okay thank you, I understand now
DrBob222
Yes, the equation tells you that 1 mol NaHCO3 produces 1 mol CO2; therefore, mols NaHCO3 = mols CO2.