TBR General Chemistry Question

[happyfellow]

Hi everyone, I'm looking at TBR Gen Chem Question in the book, example 4.13.

How many milliliters of 0.6M HCl are required to neutralize 3.0 grams CaCO3?

The answer gives this balanced reaction:

CaCO3 + 2 HCl --> CaCl2 + CO2 + H2O

Here's my though process. Since there are 2 molecules of HCl per molecule of CaCO3 we need to multiple 0.6M by 2 (because of the amount of acidic protons in solution) and then set that equal to .03 moles of CaCO3. We then solve for volume with the following equation:
2*.6M HCl*(x) = .03moles CaCO3

Could someone please clarify and provide an answer? Thanks!

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[JohnnyBravo]

Hi everyone, I'm looking at TBR Gen Chem Question in the book, example 4.13.

How many milliliters of 0.6M HCl are required to neutralize 3.0 grams CaCO3?

The answer gives this balanced reaction:

CaCO3 + 2 HCl --> CaCl2 + CO2 + H2O

Here's my though process. Since there are 2 molecules of HCl per molecule of CaCO3 we need to multiple 0.6M by 2 (because of the amount of acidic protons in solution) and then set that equal to .03 moles of CaCO3. We then solve for volume with the following equation:
2*.6M HCl*(x) = .03moles CaCO3

Could someone please clarify and provide an answer? Thanks!

You are on the right track.

In a standard neutralization reaction the number of moles must be equal. Consequently, you did the right thing in solving for moles.

The number of moles of H+ needed is what? .06 moles to neutralize .03 moles of CaCO3. So, if I have .6M *x=.06 moles. Then, i take .06/.6= .1 Liters. So, I need 100 mL of .6 M HCl.

Now, you're right that it takes 2 moles of HCl to neutralize one mole of CaCO3. You do NOT double the concentration of HCl as that would mean you have a 1.2 Molar. You simply double the moles of CaCO3. Then since you have the moles that you want and the concentration you can solve for volume.

bravo

 

[BerkReviewTeach]

You are on the right track.

In a standard neutralization reaction the number of moles must be equal. Consequently, you did the right thing in solving for moles.

The number of moles of H+ needed is what? .06 moles to neutralize .03 moles of CaCO3. So, if I have .6M *x=.06 moles. Then, i take .06/.6= .1 Liters. So, I need 100 mL of .6 M HCl.

Now, you're right that it takes 2 moles of HCl to neutralize one mole of CaCO3. You do NOT double the concentration of HCl as that would mean you have a 1.2 Molar. You simply double the moles of CaCO3. Then since you have the moles that you want and the concentration you can solve for volume.

bravo

Bravo! JBravo, that's eactly it and very well put.

 

[shahcoco]

Bravo! JBravo, that's eactly it and very well put.

You are on the right track.

In a standard neutralization reaction the number of moles must be equal. Consequently, you did the right thing in solving for moles.

The number of moles of H+ needed is what? .06 moles to neutralize .03 moles of CaCO3. So, if I have .6M *x=.06 moles. Then, i take .06/.6= .1 Liters. So, I need 100 mL of .6 M HCl.

Now, you're right that it takes 2 moles of HCl to neutralize one mole of CaCO3. You do NOT double the concentration of HCl as that would mean you have a 1.2 Molar. You simply double the moles of CaCO3. Then since you have the moles that you want and the concentration you can solve for volume.

bravo

How do you know you need .06 mol of HCl?

 

[sazerac]

You have .03 moles of CaCO3, but each of those CO3's needs TWO HCl's according to the balanced equation.

 

[shahcoco]

I still don't understand. Why do you multiply the number of equivalents (2) by the moles of CO3 vs. molarity of HCl?

 

[sazerac]

I still don't understand. Why do you multiply the number of equivalents (2) by the moles of CO3 vs. molarity of HCl?

While I could answer your question as asked, I get the impression you are just flailing at the problem and grabbing numbers off the page instead of solving it in a logical stepwise fashion. Here is how to solve basically every chemistry problem:

What did they give you? Grams of CaCO3. What do they want? mL of HCl soln. So we need a series of steps to convert from one to the other. Just like every other chemistry problem ever written.

First convert grams of CaCO3 to moles of CaCO3. How? Use a periodic table. Periodic tables are for converting between mass and moles.

Next, convert moles of CaCO3 to moles of HCl. How? Use the balanced equation. Balanced equations only work in moles.

Next convert moles of HCl to mL of HCl soln. How? Use the molarity. Molarity is for converting between moles of solute and liters of solution. Now you have your answer.

The whole problem can be solved as: 3.0g CaCO3 x (1 mol CaCO3 / 100g CaCO3) x (2 mol HCl / 1 mol CaCO3) x (1 L HCl soln / 0.6 mol HCl) x (1000 mL / 1 L).

Every unit cancels every other unit, and you have now converted grams of CaCO3 to mL of HCl soln.

 

[brood910]

Or, the easiest way to think about this is that
in mv = mv, just multiply the # of equivalents to the side where the # moles is given.