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Chemistry as level help

[h="1"]1.24g of calcium carbonate is reacted with 15 ml of 1M hydrochloric acid. How many dm3 of gas are formed? How many atoms are there in this gas?[/h]
I don't get what to do.. I worked out the balanced formula of:
CACO3 + 2HCL -> CACL2 +CO2 + H2O

I got the moles of calcium carbonate as 62/5005 moles, and the moles of hydrochloric acid as 0.015 moles. However, I'm confused what I do with these, normally you compare the ratio of moles, but which one do I compare C02 with, Calcium carbonate or hydrochloric acid??

Reply 1

mattcandy3

Work out number of moles of CaCO3 using moles= mass/Mr so 1.24/ Mr of CaCO3, then the ratio is 1:1 for CaCO3 and CO2 so to work out how many dm3 gas is formed do number of moles x 24, if you need more help just ask

Reply 2

MrB24

But what about the moles of HCL why do you just ignore that??

Reply 3

Annaaa123

If the HCl is in excess you ignore it, you always use the limiting reactant for mole ratios :smile:

The limiting reactant is the one which has the lowest number of moles, if you use the HCl, your answer will be much too high. Hope this helps! :smile:

Thankss

Original post by Annaaa123

The limiting reactant is the one which has the lowest number of moles

That's wrong. You can't ignore molar ratio as given by the reaction stoichiometry. Actually in this question CaCO₃ is the limiting reagent and your advice is off.

Reply 6

charco

Study Forum Helper

Original post by Borek

That's wrong. You can't ignore molar ratio as given by the reaction stoichiometry. Actually in this question CaCO₃ is the limiting reagent and your advice is off.

I don't know who negged Borek's post, but he is quite correct.

CaCO₃ + 2HCl --> products

means that every mol of calcium carbonate needs two moles of hydrochloric acid.

mol calcium carbonate = 1.24/100 = 0.0124 mol
mol hydrochloric acid = 0.015 mol

Hence for all of the calcium carbonate to react you would need 0.0124 * 2 = 0.0248 mol of HCl

you only have 0.015 mol HCl so it gets used up BEFORE all of the calcium carbonate can react.

This makes the HCl the limiting reagent, and therefore it, not the calcium carbonate, determines the moles of product formed.

0.015 mol HCl produce half the moles of carbon dioxide (from the equation stoichiometry) = 0.0075 mol

If you assume RTP this makes 0.0075 * 24 = 1.8 dm³

Reply 7

Annaaa123

Original post by Borek

That's wrong. You can't ignore molar ratio as given by the reaction stoichiometry. Actually in this question CaCO₃ is the limiting reagent and your advice is off.

Oh goodness, yes, I completely looked past the equation! Sorry! :colondollar:

Reply 8

daniela123456789

so what is the actual answer please

Reply 9

Igorzycho

Original post by charco

If you assume RTP this makes 0.0075 * 24 = 1.8 dm³

0.0075 * 24 = 0.18 dm^3

Reply 10

charco

Study Forum Helper

Original post by Igorzycho

0.0075 * 24 = 0.18 dm^3

well spotted.

I hope I have improved my maths skills since 3rd October 2013

Reply 11

Igorzycho

hehe no problem :smile:

i) I got 0.18 dm^3
ii) 1.35*10^22 atoms (however I don't know if it is correct - can someone confirm?)

(edited 6 years ago)

Reply 12

charco

Study Forum Helper

Original post by Igorzycho

hehe no problem :smile:

i) I got 0.18 dm^3
ii) 1.35*10^22 atoms (however I don't know if it is correct - can someone confirm?)

You've got the moles of molecules = 0.0075 mol

each molecule contains 3 atoms = 3 x 0.0075

and each mol contains 6 x 10²³ particles = 3 x 0.0075 x 6 x 10²³ = 1.35 x 10²²

.. but be careful!

Someone may correct it in five years time ...