I really hate AP chemistry, so i need ur help!?

Question

ok here's the question:

Some pure calcium carbonate was added to 145 mL of 1.00 mol/L HCl. When the reaction had finished, the solution was acidic. 25.0 mL of .500 mol/ L Na2CO3 was required to neutralize the excess acid. What mass of calcium carbonate was used?

CaCO3 (s) + 2HCl (aq) => CaCl2 (aq) + CO2 (g) + H2O (l)

Na2CO3 (s) + 2HCl (aq) => 2NaCl (aq) +CO2 (g) +H2O (l)

thnx a bunch

Answer 1

If u hate AP chem u shouldnt have taken it, this is all ur fault, u brought it on urself. and do ur own hw

Answer 2

You started with 0.145 moles of HCl (liters x Molarity = moles; 0.145 x 1.00 M = 0.145 moles)

25.0 mL of 0.500 MNa2CO3 was required to neutralize the left-over HCl

0.025 L x 0.500 M = 0.0125 moles Na2CO3

Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O

0.0125 mole Na2CO3 will neutralize 0.0250 mole HCl (2x0.0125)

0.145 mole HCl to start

0.025 mole HCl left

0.120 mole HCl reacted with CaCO3

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

0.120 mole HCl will react with 0.060 mole CaCO3 (0.120/2)

0.060 mole CaCO3 x 100 g/mole = 6.0 g CaCO3

Answer 3

It's really just a big long molarity and stoichiometry problem.  Breathe and take this one step at a time.  The trick is to plan strategically and solve.

A slight rephrase of the problem:

Someone used some amount of CaCO3 and a known amount of HCl for a reaction.  There was HCl left over, so they used some amount of Na2CO3 to get rid of all of the extra.

My plan of attack:

1. Find out how much HCl you started with.

2. Find out how much HCl you had extra.

3. Subtract #2 from #1 to figure out how much HCl you used.

4. Figure out how much CaCO3 would react with the amount in #3.

Step by step:

1. Figure out amount of HCl you start with using the volume and concentration information you are given.

145 mL *( 1 L/ 1000 mL) = 0.145 L

0.145 L * (1.00 mol HCl/ L) = 0.145 mol HCl

You start out with 0.145 mol HCl.

2. To figure out how much HCl you have extra takes two steps.  You need to first figure out how much Na2CO3 you have.  Then you need to figure out how much HCl you will use to react with it.  Remember, whatever HCl you have extra is used to react with the Na2CO3.

a. Figure out amount of Na2CO3 available

25.0 mL (1 L / 1000 mL) = 0.025 L

0.0250 L (0.500 mol Na2CO3/ L) = 0.0125 mol Na2CO3

b.  Figure out how many moles of HCl was used was used to react with the amount of Na2CO3 available.  You're using the mole ratio from the balanced equation here (2 HCl: 1 Na2CO3).

0.0125 mol Na2CO3 * (2 HCl/ 1 Na2CO3) = 0.0250 mol HCl.

Now you know that you have 0.0250 mol HCl left over.

3. So, you start with 0.145 mol HCl and you have 0.0250 mol HCl left over at the end of the first reaction.  That means that you used up 0.120 mol HCl in the first reaction.

4. For every 2 HCl used, 1 CaCO3 is required (again, mole ratio).  So, from that reasoning:

0.120 mol HCl * (1 CaCO3/ 2 HCl) = 0.0600 mol CaCO3.  

Since the question asks for mass, you would convert 0.0600 mol CaCO3 into grams:

0.0600 mol CaCO3 * (100.09 g/ mol) = 6.01 g CaCO3

Note: 100.9 g/mol is molar mass of CaCO3: 40.08 g + 12.01 g  + 3(16.00g).  

That's it.