Solid calcium carbonate, CaCO3(s), reacts with HCl(aq) to...

Solid calcium carbonate, $\mathrm{CaCO}_{3}(\mathrm{s}),$ reacts with $\mathrm{HCl}(\mathrm{aq})$ to form $\mathrm{H}_{2} \mathrm{O}, \mathrm{CaCl}_{2}(\mathrm{aq}),$ and $\mathrm{CO}_{2}(\mathrm{g}) .$ If a $45.0 \mathrm{g}$ sample of $\mathrm{CaCO}_{3}(\mathrm{s})$ is added to $1.25 \mathrm{L}$ of $\mathrm{HCl}(\mathrm{aq})$ that is $25.7 \% \mathrm{HCl}$ by mass $(d=1.13 \mathrm{g} / \mathrm{mL})$ what will be the molarity of HCl in the solution after the reaction is completed? Assume that the solution volume remains constant.

Key Concepts

Density and Volume Conversion

Density connects the mass and volume of a substance, and this relationship is used to convert between the two. For solution calculations, knowing the density allows you to determine the total mass of the solution from its volume, which, when combined with percent composition, helps in calculating the mass and moles of the solute.

Stoichiometry of Chemical Reactions

This concept involves writing and interpreting balanced chemical equations to determine the exact ratios in which reactants combine and products form. It is essential for relating the moles of one substance to the moles of another, enabling calculations such as determining how many moles of HCl are required or remain after reacting with CaCO?.

Limiting Reactant

The limiting reactant is the substance in a chemical reaction that is entirely consumed first, limiting the amount of product produced and affecting the amount of any excess reactant remaining. Understanding which reagent is limiting is crucial for accurately calculating the post-reaction concentrations of the remaining reactants.

Percentage Composition

This key concept involves using percent by mass to determine the mass of a specific component in a mixture or solution. In the context of the problem, it allows for the calculation of the mass of HCl present in a solution given its percentage composition by mass.

Molarity Calculation

Molarity is defined as the number of moles of solute per liter of solution. This concept is used to determine the concentration of a chemical species in solution after accounting for the reaction, making it necessary for assessing the remaining concentration of HCl after the reaction has occurred.

Transcript

00:01 Here we have a reaction that consumes hcl to produce calcium chloride.

00:06 We're given essentially a starting concentration of hcl, and we're given a massive calcium carbonate that's consumed and asked what's the final concentration of the hcl solution.

00:17 So let's start by how much hcl do we have.

00:22 We have 1 .25 liters, and its density is 1 .13 grams per mill, but only only.

00:30 25 .7 % by mass of that solution is actually hcl.

00:36 So let's work out the mass of the solution.

00:41 For starters, and remember, density is mass per unit volume.

00:48 So the mass of the solution is going to be that density, 1 .13 grams per milliliter.

00:57 Now i'm going to have to times that by 1 ,000 to put it into milliliters.

01:03 So that's 1 ,250 milliliters.

01:11 So the mass of the solution is 1 ,420 .5 grams.

01:23 But to get the actual hcl, we're going to have to multiply by 25 .7%, divide by 100 to get that into a decimal.

01:35 So that's .257.

01:39 So my massive hcl in my starting solution is 363 .01 grams.

01:57 Okay.

01:58 Now, how many moles of hcl is that? our molecular weight of hcl is 36 .458.

02:17 And so we're starting with, remember, molecular weight is mass divided by number of numbers, grams divided by 36 .45 grams per mole.

02:55 So we started with 9 .9 .6569 moles of hcl.

03:10 Okay, now let's look at our calcium chloride and see how much of that was consumed...

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