Question

You have 12.5 mL of a 0.20 M solution of sulfuric acid that you want to neutralize with a solution of
sodium bicarbonate according to the following neutralization reaction:
H2S04 (ag) + 2 NaHCO3 (ag) -> 2 H20 (1) + 2 CO2 (g) + Na2S04 (ag)
a. Calculate the moles of sulfuric acid in 12.5 mL of a 0.20 M H2S04 solution (5 pts).
MY ANSWER
12.5 mL. = 0.0125 L. So then 0.20 x0.0125 = 0.0025 moles of sulfuric acid
b. Calculate the moles of sodium bicarbonate (NaHCO3) that will react with the amount of
sulfuric acid that you calculate in your part A (4 pts).
MY ANSWER
Since 1 mol of H2S04 = 2 mol NaHCO3. 0.0025 x2 = 0.005 moles of sodium
bicarbonate
C.
Using your answer to par B above, calculate the milliliters of 2.0 M NaHCO3 needed to
neutralize the 12.5 mL of a 0.20 M sulfuric acid (5 pts)
I NEED HELP ON THIS
d.
Calculate the number of liters of Co2 gas that will be produced by completely neautralizing 12.5
mL of a 0.20 M H2S04 with NaHCO3 at 25 C and 0.95 atm. Hints: Use the answer you
calculated in part B (moles of NaHCO3 = moles of CO2), and the ideal gas law equation (6 pts).
HELP ON THIS TOO

Answer 1

First, let me be picky. You typed a 0 for O in H2SO4 and H2O and Na2SO4. Also, you typed a g for a q to show aqueous solution. If you did this on an exam for me I would NOT have taken off points but I would have corrected it this way. I also see a 1 for an l to show liquid H2O.
H2SO4 (aq) + 2 NaHCO3 (aq) -> 2 H2O (l) + 2 CO2 (g) + Na2SO4 (aq)
Your answer to a and correct. Ditto for b. Good work.
c. You need 0.005 mols NaHCO3 that were calculated in part B. You calculated that with mols = M x L. That will get you almost anywhere you want to go. In part C you know moles = 0.005. You know M = 0.20. Solve for L and convert to mL.
d. Look at the hint: moles NaHCO3 = mols CO2. Use PV = nRT. You know n = mols NaHCO3 = moles CO2; I assume you know R, P is given and T is given in degrees C. Remember to convert temperature from C to Kelvin.
K = 273 + C = ?
Post your work if you get stuck.