[Runner19]

I'm extremely confused on how to solve this problem even after reading my textbook. I couldn't find a problem in the textbook similar to this one either. The problem reads as the following:

5.27 g of calcium carbonate reacts with 10.3 mL of 0.10 M hydrochloric acid. What mass of carbon dioxide could be produced if all of at least one reactant is used completely?

So i started by balancing the equation:

CaCO3 + 2HCl  CaCl2 + H2O + CO2

Next, I converted 5.27 g of CaCO3 to 0.0526 moles of CaCO3.
I also used the formula n = MV  to get n = 0.0103 mL * 0.10 M = 0.00103

Beyond this, I'm lost. Assuming (hopefully) that my calculations above were correct, could someone explain to me the next couple steps I'm to take to figure out this problem?

[Borek]

What you did so far is OK.

Do you know the concept of the limiting reagent?

[Arkcon]

OK, we've had one of these recently.  Lets see if you can work it out.

I'm extremely confused on how to solve this problem even after reading my textbook. I couldn't find a problem in the textbook similar to this one either.

That seems unlikely to me.  Just sharing that point with you.  We can still work on it together.

The problem reads as the following:

5.27 g of calcium carbonate reacts with 10.3 mL of 0.10 M hydrochloric acid. What mass of carbon dioxide could be produced if all of at least one reactant is used completely?

Got it.

So i started by balancing the equation:

CaCO3 + 2HCl  CaCl2 + H2O + CO2

Good.

Next, I converted 5.27 g of CaCO3 to 0.0526 moles of CaCO3.

Good.

I also used the formula n = MV  to get n = 0.0103 mL * 0.10 M = 0.00103

Possibly just as good, but keeping the units could come in handy.

Beyond this, I'm lost. Assuming (hopefully) that my calculations above were correct, could someone explain to me the next couple steps I'm to take to figure out this problem?

So you balanced the equation.  Its an equation, so it also has units.  What are they?  We don't go the the shelf, pick up a molecule of CaCO₃, then stick on two HCl's.  Hint: you knew to convert grams and concentration per volume into moles.  Why did you do this?

[Runner19]

Ok. Am I right to divide 0.526 moles of CaCO3 by itself? The balanced equation has 1 mole of CaCO3, so I can divide the 0.00103 moles of HCl by 0.526, which gives me around 0.19. Since it's less than 2, the HCl would by me limiting reagent, right?

[Borek]

Ok. Am I right to divide 0.526 moles of CaCO3 by itself?

Not that we can stop you from doing it, but I don't see how it is going to help.

The balanced equation has 1 mole of CaCO3, so I can divide the 0.00103 moles of HCl by 0.526, which gives me around 0.19. Since it's less than 2, the HCl would by me limiting reagent, right?

That looks much more reasonable, although I don't see a logical connection between the first statement (1 mole of CaCO₃) and the rest of the phrase. But yes, HCl is the limiting reagent, and it is an obvious conclusion of the fact [itex]\frac{0.00103}{0.526}< 2[/itex].

[Vidya]

CaCO3 + 2HCl  CaCl2 + H2O + CO2

Next, I converted 5.27 g of CaCO3 to 0.0526 moles of CaCO3.
I also used the formula n = MV  to get n = 0.0103 mL * 0.10 M = 0.00103

Do not think very hard and make a Simple chemistry problem tough.
As indicated by Borek it is a problem of limiting reactant .Use balanced equation and convert moles of CaCO3 into moles CO2 and moles of HCl into moles of CO2.Check which one is giving you minimum moles of CO2 and that is limiting reactant. Convert these moles of CO2 into grams .