You place 2.56 g of CaCO3 in a beaker containing 250 . mL of 0.125 M HCl. When the reaction has

You place 2.56 g of CaCO3 in a beaker containing 250 . mL of...

You place $2.56 \mathrm{g}$ of $\mathrm{CaCO}_{3}$ in a beaker containing $250 .$ mL of $0.125 \mathrm{M}$ HCl. When the reaction has ceased, does any calcium carbonate remain? What mass of $\mathrm{CaCl}_{2}$ can be produced? $$\mathrm{CaCO}_{3}(\mathrm{s})+2 \mathrm{HCl}(\mathrm{aq}) \rightarrow \mathrm{CaCl}_{2}(\mathrm{aq})+\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell)$$

Key Concepts

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is completely consumed first, thus stopping further reaction and determining the maximum amount of product that can be formed. Understanding which reactant is limiting allows one to calculate the theoretical yield of a reaction accurately.

Stoichiometry

Stoichiometry involves the quantitative relationships between reactants and products in a balanced chemical reaction. It uses mole ratios to convert between amounts of substances, enabling the calculation of how much product can be formed from given reactant quantities.

Molarity and Volume

Molarity is a measure of a solute’s concentration in a solution, defined as the number of moles of solute per liter of solution. By combining molarity with the volume of the solution, one can determine the number of moles of a reactant present in a given solution.

Theoretical Yield

The theoretical yield is the maximum amount of product that can be produced from a given amount of reactants, based on stoichiometric calculations. It assumes complete reaction of the limiting reactant with no losses or side reactions.

Mass-Mole Conversion

Mass-mole conversion is the process of converting between the mass of a substance and the number of moles using the molar mass. This conversion is essential in stoichiometric calculations to relate experimental masses to the stoichiometric coefficients in the chemical equation.

Transcript

00:01 Okay, in this problem, we are given the reaction between calcium carbonate and hydrochloric acid.

00:07 I'm doing a voiceover on this, so there'll be some downtime.

00:10 We're given the following chemical equation.

00:13 Calcium carbonate reacts with hydrochloric acid, and our products are calcium chloride, carbon dioxide, and h2o.

00:20 We are given 2 .56 grams of calcium carbonate.

00:26 And we're given 250 milliliters of 0 .125 million liters of 0 .125.

00:32 Molar hydrochloric acid.

00:38 We can see from looking at this, these pieces of information, that we are dealing with the limiting reactant problem.

00:48 We are going to be asked to determine which reactant is limiting, how much excess reactant we have, and how much calcium chloride could theoretically be produced.

01:02 We have a beaker containing the acid.

01:05 We add the calcium carbonate.

01:07 Reaction occurs.

01:09 One of two things will happen.

01:11 We'll either run out of calcium carbonate and have excess hydrochloric acid, or we'll run out of acid and have excess calcium carbonate.

01:21 I've already mentioned what the tasks are.

01:23 We're going to find which reactant is excess, or excuse me, which reactant is limiting, and how much excess remains and how much cacl2 is produced.

02:01 I always have downtime when i'm doing these voicemowers.

02:13 Okay, for the calcium carbonate, we're going to do this conversion.

02:22 Mass of a, that's your given, to moles of a, to moles of b, to mass of b.

02:31 And that means massive calcium carbonate to moles of calcium carbonate.

02:34 Then we're going to do a mole conversion for calcium carbonate to calcium chloride.

02:39 And then we'll convert to massive calcium chloride.

02:42 There's other ways to do this, but with the ease of punching.

02:47 Numbers into a calculator, i'd just as soon punch them up and be done.

02:52 The molar mass of calcium carbonate is 100 .09 grams per mole, and the molar mass of calcium chloride is 110 .98 grams per mole.

03:09 Let's do this mass mass, mass, stochymetric calculation.

03:15 We were given 2 .56 grams of calcium carbonate.

03:25 Okay, i mess around with that a little bit now, so i'm just going to check and make sure that's still sorry about that.

03:39 Keep erasing.

03:41 Okay, now i'm going to raise one more time, and then 2 .56 grams of calcium carbonate.

03:50 Now we're going to use the molar mass of calcium carbonate of 100 .09 grams and one mole in the numerator.

04:04 Our next step will be used to use the mole ratio.

04:08 The only coefficient we had was for hydrochloric acid, and that was one...

Please give Ace some feedback