(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)?
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). :-)
for c. I got 0.1 mol HCl
Can you confirm if it's right or not
c. To determine the number of moles of hydrochloric acid used, we need to calculate the difference in the amount of sodium hydrogen carbonate before and after the reaction.
First, we convert the given masses of sodium hydrogen carbonate into moles using the molar mass of NaHCO3. The molar mass of NaHCO3 is calculated by adding up the atomic masses of each element in the compound:
Na (22.99 g/mol) + H (1.01 g/mol) + C (12.01 g/mol) + O (16.00 g/mol) + 3 O (16.00 g/mol) = 84.01 g/mol
To convert the grams of sodium hydrogen carbonate into moles, we divide the mass by the molar mass:
Initial moles of NaHCO3 = 5.5 g / 84.01 g/mol
Final moles of NaHCO3 = 1.3 g / 84.01 g/mol
To find the moles of hydrochloric acid used, we need to consider the stoichiometry of the reaction. From the balanced equation, we see that one mole of NaHCO3 reacts with 2 moles of HCl:
2 moles HCl = (Initial moles of NaHCO3 - Final moles of NaHCO3) * 2
Plug in the values:
2 moles HCl = (5.5 g / 84.01 g/mol - 1.3 g / 84.01 g/mol) * 2
Simplify and calculate:
2 moles HCl = (4.65 - 0.03) * 2
2 moles HCl = 4.62 * 2
2 moles HCl = 9.24 moles of HCl
Therefore, approximately 9.24 moles of hydrochloric acid were used in the reaction.
d. To determine the volume of gas produced at STP by the reaction, we need to use the ideal gas law. The number of moles of carbon dioxide (CO2) produced in the reaction is equal to the number of moles of hydrochloric acid used.
Using the concept that one mole of any gas occupies 22.4 L at STP, we can calculate the volume of CO2 produced:
Volume of CO2 = number of moles of CO2 * 22.4 L/mol
Since the number of moles of CO2 is equal to the number of moles of HCl used:
Volume of CO2 = 9.24 moles * 22.4 L/mol
Volume of CO2 = approximately 206.98 L
Therefore, approximately 206.98 liters of gas at STP would be produced by the reaction described in part c).