as level chemistry exam question help!

the equation for the reaction between magnesium carbonate and hydrochloric acid is given below,
MgCO3 + 2HCl = MgCl2 +H2O +CO2

when 75 cm3 of 5.00 mol dm-3 hydrochloric acid were added to 1.25g of impure MgCO3 some acid was left unreacted. This unreacted acid required 21.6cm3 of a 0.500 mol dm-3 solution of sodium hydroxide for complete reaction.

a) show that the number of moles of HCl which reacted with the MgCO3 in the sample was 0.0267.

b) Calculate the number of moles and the mass of MgCO3 in the sample, and hence deduce the percentage of MgCO3 in the sample.

Reply 1

shengoc

12

08jimbojames

the equation for the reaction between magnesium carbonate and hydrochloric acid is given below,
MgCO3 + 2HCl = MgCl2 +H2O +CO2

when 75 cm3 of 5.00 mol dm-3 hydrochloric acid were added to 1.25g of impure MgCO3 some acid was left unreacted. This unreacted acid required 21.6cm3 of a 0.500 mol dm-3 solution of sodium hydroxide for complete reaction.

a) show that the number of moles of HCl which reacted with the MgCO3 in the sample was 0.0267.

b) Calculate the number of moles and the mass of MgCO3 in the sample, and hence deduce the percentage of MgCO3 in the sample.

Hmm, probably you have copied the question wrongly here, it works if the amount of HCl used at first is 7.5 instead of 75 cm^3, but anyway, you have the excess acid reacted with NaOH.

The stoichiometry if you make a balanced eqn, then the mole of excess acid = mole of NaOH reacted = 0.5 x 21.6 /1000 = 0.0108 mol

Now you have the total amount of HCl added at first = 7.5/1000 x5 = 0.0375

Hence, the amount of HCl actually reacted = Total - excess = 0.0375 - 0.0108 = 0.0267 mol(as required)

Reply 2

imtired

08jimbojames

the equation for the reaction between magnesium carbonate and hydrochloric acid is given below,
MgCO3 + 2HCl = MgCl2 +H2O +CO2

when 75 cm3 of 5.00 mol dm-3 hydrochloric acid were added to 1.25g of impure MgCO3 some acid was left unreacted. This unreacted acid required 21.6cm3 of a 0.500 mol dm-3 solution of sodium hydroxide for complete reaction.

a) show that the number of moles of HCl which reacted with the MgCO3 in the sample was 0.0267.

b) Calculate the number of moles and the mass of MgCO3 in the sample, and hence deduce the percentage of MgCO3 in the sample.

i think you mean 0.5moldm-3 there, i'll assume you do.

you have 75cm3 of 0.5moldm-3 solution at first, you need to work out the number of moles of that.

some of it reacted, but some didn't. you want to find out how much reacted, but you've got how much didn't react. the amount that didn't react was neutralized by 21.6cm3 of a 0.5moldm-3 solution. from this you can work out how much reacted, because it will be the same number of moles.

take this away from the total amount of hcl you had initially and you've got the number of moles of hcl that reacted with the mgco3.

the equation you need is moles = concentration x volume (in dm3, so if you're given a volume in cm3 you need to divide by 1000).

Reply 3

shengoc

12

08jimbojames

the equation for the reaction between magnesium carbonate and hydrochloric acid is given below,
MgCO3 + 2HCl = MgCl2 +H2O +CO2

when 75 cm3 of 5.00 mol dm-3 hydrochloric acid were added to 1.25g of impure MgCO3 some acid was left unreacted. This unreacted acid required 21.6cm3 of a 0.500 mol dm-3 solution of sodium hydroxide for complete reaction.

a) show that the number of moles of HCl which reacted with the MgCO3 in the sample was 0.0267.

b) Calculate the number of moles and the mass of MgCO3 in the sample, and hence deduce the percentage of MgCO3 in the sample.

Now for the second part, mole of MgCO3 actually reacted = 1/2 x mol of HCl reacted = 1/2 x 0.0267 =

mole = mass/rmm, so mass of MgCO3 = RMM OF MgCO3 x 1/2 x 0.0267 =

Hence percentage of MgCO3 = Actual mass of MgCO3 / 1.25 x 100 % = (you work out yourself)

Hope that helps.

Reply 4

imtired

shengoc

Hmm, probably you have copied the question wrongly here, it works if the amount of HCl used at first is 7.5 instead of 75 cm^3, but anyway, you have the excess acid reacted with NaOH.

nobody replies for an hour and when i decide to someone beats me by one minute!

i assumed the concentration was 10 times smaller but numerically you get the same answer.

Reply 5

shengoc

12

imtired

nobody replies for an hour and when i decide to someone beats me by one minute!

i assumed the concentration was 10 times smaller but numerically you get the same answer.

Yeah, i just got online, hmm....the values do seem odd, but either way, it would still work out, thanks to that "show" first question.

Reply 6

EasyTiger

3

So the answer is 90% then?

Reply 7

shengoc

12

EasyTiger

So the answer is 90% then?

If you do the right calculations, you should get the right answer. i am not doing the calculations, i am just telling you the steps....the answer is not important after all, if you know how to do it, you can work out any questions,ok?

Reply 8

sciencegeek

shengoc

Now for the second part, mole of MgCO3 actually reacted = 1/2 x mol of HCl reacted = 1/2 x 0.0267 =

mole = mass/rmm, so mass of MgCO3 = RMM OF MgCO3 x 1/2 x 0.0267 =

Hence percentage of MgCO3 = Actual mass of MgCO3 / 1.25 x 100 % = (you work out yourself)

Hope that helps.

RMM OF MgCO3 x 1/2 x 0.0267

do you times by 1/2 because of the concentration or that there is 1MgCO3 to 2HCL...half as much MgCO3 compared to HCL?

many thanks :smile:

PS sorry if it's a stupid question and the answers obvious, ive just taken up chemistry so im still a bit of a newb, lol

Reply 9

NatsP29

15

why do you divide by 1.25 ?

Reply 10

Callum787

1

Original post by shengoc

Hmm, probably you have copied the question wrongly here, it works if the amount of HCl used at first is 7.5 instead of 75 cm^3, but anyway, you have the excess acid reacted with NaOH.

The stoichiometry if you make a balanced eqn, then the mole of excess acid = mole of NaOH reacted = 0.5 x 21.6 /1000 = 0.0108 mol

Now you have the total amount of HCl added at first = 7.5/1000 x5 = 0.0375

Hence, the amount of HCl actually reacted = Total - excess = 0.0375 - 0.0108 = 0.0267 mol(as required)