(Can you check my work for a. and b.? I'm confused on how to do c. and d.)

When hydrochloric acid is dropped onto a sample of sodium hydrogen carbonate, NaHCO3, the many bubbles that form are a clear indication that a vigorous chemical reaction occurs. Answer the following questions about the reaction between hydrochloric acid and sodium hydrogen carbonate (baking soda).
a. Write out the complete chemical reaction. Include reactants, products, and the state of matter for each reactant and product.
NaHCO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)

b. Identify the type of reaction.
Acid- base

c. In an experiment, hydrochloric acid is dropped onto a 5.5 gram sample of sodium hydrogen carbonate. After all of the acid has been added, 1.3 grams of sodium hydrogen carbonate remains. how many moles of hydrochloric acid were used?

d. At standard temperature and pressure (STP) one mole of any gas occupies 22.4L. What volume of gas at STP would be produced by the reaction described in c)?

2 answers

for c. I got 0.1 mol HCl
Can you confirm if it's right or not
Your answer to a is correct. As for b, I think your answer is OK but I think the chemistry community should get their act together BECAUSE authors just don't agree. For example, most authors list five BASIC types of reactions and the acid base (what I prefer to call neutralization reactions) reaction is not one of them. Another author lists SIX basic reactions and NEITHER author lists redox reactions as a type of reaction. Anyway, I would go with acid-base as a correct answer.

c. This returns to several of your questions about equations and convert mols of one material to moles of another.
NaHCO3(aq) + 2HCl(aq) → 2NaCl(aq) + CO2(g) + H2O(l)
mols NaHCO3 initially = 5.5/84 = 0.0655
mols NaHCO3 final = 1.3/84 = 0.0155
mols NaHCO3 used = 0.0655 - 0.0155 = 0.05
So you want to convert 0.05 mols NaHCO3 used to mols HCl used.
0.05 mols NaHCO3 x (2 mol HCl/1 mol NaHCO3) = 0.05 x 2/1 = 0.1 mol HCl used.

d. You had 0.05 mols NaHCO3 used. You want to convert that to mols CO2 released in the reaction. That's 0.05 mols NaHCO3 x (1 mol CO2/1 mol NaHCO3) = 0.05 x (1/1) = 0.05 moles CO2.
@ STP this will occupy 0.05 mols x 22.5 L/mol = ? L. As an aside, note that the units of mols x L/mol = L and that's the unit you want so you know to multiply by 22.4 L/mol and not (1 mol/22.4 L). :-)