Solve Stoichiometry Problems with HCl and Marble

[DB]

A little stoichiometry...
Here's all that was given:

A container of HCl spilled on a marble table surface. The marble contains calcium carbonate, which reacts with the acid to produce calcium chloride, carbon dioxide and water.

a) If 500ml of 0.2 M acid spilled, what mass of calcium carbonate reacted?

So I wrote out the (combustion is it?) equation.
CaCO_3 + 2HCl\rightarrow CaCl_2 + CO_2 + H_2O

Then molar mass:
100g(CaCO_3) + 72g(2HCl)\rightarrow 110g(CaCl_2) + 44g(CO_2) +18g( H_2O)

So now I know that 72g of HCl is 2 mol (since 2HCl) so 0.2 mol = 7.2g
But know I'm stuck...a push in the right direction?

Also.
b)How many moles of CO_2 were formed during this reaction?
1 mole
c)How many molecules of water were formed during this reaction?
~6.022*10^23
I don't think there's any problems there.

Thanks in advance

 

[dextercioby]

Yes,there are problems everywhere.So the first.U've computed 0.2 mols of HCl.How many mols of CaCO_{3} will react with it...?

Daniel.

 

[saltydog]

A little stoichiometry...
Here's all that was given:

A container of HCl spilled on a marble table surface. The marble contains calcium carbonate, which reacts with the acid to produce calcium chloride, carbon dioxide and water.

a) If 500ml of 0.2 M acid spilled, what mass of calcium carbonate reacted?

So I wrote out the (combustion is it?) equation.
CaCO_3 + 2HCl\rightarrow CaCl_2 + CO_2 + H_2O

Then molar mass:
100g(CaCO_3) + 72g(2HCl)\rightarrow 110g(CaCl_2) + 44g(CO_2) +18g( H_2O)

So now I know that 72g of HCl is 2 mol (since 2HCl) so 0.2 mol = 7.2g
But know I'm stuck...a push in the right direction?

Also.
b)How many moles of CO_2 were formed during this reaction?
1 mole
c)How many molecules of water were formed during this reaction?
~6.022*10^23
I don't think there's any problems there.

Thanks in advance

According to the balanced equation, for every mole of HCl reacting, one-half of a mole of CaCO_3 reacts. So how many moles of HCl are in 500ml of a 0.2 molar solution? Suppose that's x number of moles. Now, from what I said above, one-half of that number of moles is the number of moles of CaCO_3 reacting. Same dif for CO_2 right (half as much) and water too for that matter.


Now, isn't there Avagadro's number of molecules in one mole? So using the infor above (the number of moles of water produced), figure out how many molecules of water.

 

[Gokul43201]

DB,

0.2 M does not mean 0.2 moles. It refers to a solution of strength 0.2 moles per liter. So, how many moles will there be in 500 mL of this solution ?

 

[dextercioby]

Ooooooops,didn't see the big M,sorry.

Daniel.

 

[The Bob]

a) If 500ml of 0.2 M acid spilled, what mass of calcium carbonate reacted?

So I wrote out the (combustion is it?) equation.
CaCO_3 + 2HCl\rightarrow CaCl_2 + CO_2 + H_2O

Lets start here. You have the concentration and the volume. You can use these to see what the amount of moles is.

The equation is: Concentration (mol dm^{-3}) = \frac{Moles (moles)}{Volume (dm^3)}

Use that to find the number of moles of acid that have been spilled. You can then use this to find the mass: Moles (moles) = \frac{Mass (grams)}{Molar Mass (g mol^{-1})}

You then have the mass of the HCl. Use this you can find the mass of CaCl₂ in the normal way, find ratios and molar masses and see what is produced.

Get this and then I will help with the rest.

The Bob (2004 ©)

 

[DB]

Ok so, 0.1 moles reacts with CaCO_3 in a ratio of 2:1? So, half of 0.1 mol is 0.05 mole which is 5g of CaCO_3

for b and c, i think you told me that the answers are 0.5 mol and N_A/2, but i don't really understand why I'm dividing by 2...

 

[The Bob]

I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl₂ is 111 g mol^-1.

The Bob (2004 ©)

 

[saltydog]

Ok so, 0.1 moles reacts with CaCO_3 in a ratio of 2:1? So, half of 0.1 mol is 0.05 mole which is 5g of CaCO_3

Ok.

for b and c, i think you told me that the answers are 0.5 mol and N_A/2, but i don't really understand why I'm dividing by 2...

No. I think you mean 0.05 moles right? If 0.1 moles of HCl are reacting and the ratio is 2/1 then that means 0.05 moles of water and carbon dioxide are produced right? And if one mole has Avagadros number then one half of a mole would have one half of Avagadros number, one tenth of a mole would have one-tenth of Avagadros number and so if you have 0.05 moles of water then that means you have 0.05 of Avagadros's number of molecules right?

 

[DB]

"I got 55.5g of Calcium Chloride. More accurate molar mass of CaCl2 is 111 g mol-1.

The Bob (2004 ©)"
calium chloride? for a? the question is asking calcium carbonate

 

[The Bob]

calium chloride? for a? the question is asking calcium carbonate

I apologise. In that case I get 50g of Calcium Carbonate.

The Bob (2004 ©)

 

[DB]

Ok.



No. I think you mean 0.05 moles right? If 0.1 moles of HCl are reacting and the ratio is 2/1 then that means 0.05 moles of water and carbon dioxide are produced right? And if one mole has Avagadros number then one half of a mole would have one half of Avagadros number, one tenth of a mole would have one-tenth of Avagadros number and so if you have 0.05 moles of water then that means you have 0.05 of Avagadros's number of molecules right?

Naww lol i meant 0.5. I am confused. 0.1 moles of acid are reacting we know that.
So then 0.1 mole of both CO2 and water are...? I am sorry if I am being anoying I am just really confused.

 

[DB]

I apologise. In that case I get 50g of Calcium Carbonate.

The Bob (2004 ©)

no prob, just i got 5 g and saltydog said it was ok...

 

[The Bob]

Here is what I did.

The amount of moles of the spilled Hydrochloric Acid is 0.2 mol dm^-3 x 0.5 dm³ = 0.4 moles.

Molar Mas of HCl = 36.5 g mol^-1

0.4 moles x 36.5 g mol^-1 = 14.6g (now proven)

The equation you need to look at is the 2HCl + CaCO_3

HCl = 35.6 g mol^-1 x 2 moles = 73g
CaCO₃ = 100 g mol^-1 x 1 mol = 100g

Therefore HCl:CaCO₃ = 73g:100g

So 1g of HCl : 1.37g of CaCO₃

Therefore 14.6g of HCl : 1.37 x 14.6 = 20g of CaCO₃

Ok. I have changed my mind because my original answer was wrong because I get the molar mass of HCl wrong () but I am now sure it is 20g.

The Bob (2004 ©)

 

[The Bob]

The moles of Carbon Dioxide = 0.2 moles.
The moles of Water = 0.2 moles as well. I don't know about how many molecules but I personally think it might be the same.

The Bob (2004 ©)

 

[DB]

Here is what I did.

The amount of moles of the spilled Hydrochloric Acid is 0.2 mol dm^-3 x 0.5 dm³ = 0.4 moles.



The Bob (2004 ©)

Thanks bob, i see the work u've done, i just can't understand how u got 0.4 moles of acid spilled, i always get 0.1...can u explain please

 

[The Bob]

Concentration (mol dm^{-3}) = \frac{Moles (mol)}{Volume (dm^3)}

M (or the molarity) is equal to the same number in mol dm^-3 e.g. 0.2M = 0.2 mol dm^-3.

Also the equation needs the amount in dm³. You have it in ml. This can be changed straight away into cm³ e.g. 300 ml = 300 cm³. Then you need it in dm³. As it is a cubic factor then you need to change the number by a cubic number. Let me explain this better.

1 dm = 10 cm. This is worked out by multiplying the dm by 10 or by dividing the cm by 10.

1 dm² = 100 cm². This time it is worked out by multiplying the dm by 10² (100) or by dividing the cm by 10² (100).

Therefore 1 dm³ = 1000 cm³ multiplying/dividing by 10³ (1000). e.g. 300 ml = 300 cm³ = 0.3 dm³

The amount of moles of the spilled Hydrochloric Acid is 0.2M x 500 ml = 0.2 mol dm^-3 x 500 cm³ = 0.2 mol dm^-3 x 0.5 dm³ = 0.1 moles. (Oh man, I have done it again. I need to learn to check what I am doing. Sorry again ).

The Bob (2004 ©)

 

[The Bob]

I get 5g as well .[/size] Sorry everyone. I has been good practise for me though and I can see my mistakes clearer and also you can see my method clearly and rather than jumping to conclusion it can be proven (although jumping seemed to work better)[/size].

The Bob (2004 ©)

 

[The Bob]

The moles of Carbon Dioxide = 0.2 moles.
The moles of Water = 0.2 moles as well.

It is now:
The moles of Carbon Dioxide = 0.05 moles.
The moles of Water = 0.05 moles.

The Bob (2004 ©)

P.S. Sorry again but I need to sleep now. I am really quite tired. Hope what I did helped in some way or another.