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mirams
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Calcium and its compounds, have properties typical of Group 2 in the Periodic Table.
Calcium carbonate, CaCO3, reacts with acids such as nitric acid.
A student neutralised 2.68 g of CaCO3 with 2.50 mol dm–3 nitric acid, HNO3.
The equation for this reaction is shown below.
CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
(i) Determine the amount, in mol, of CaCO3 reacted.
amount = ............................................... mol
(ii) Calculate the volume, in cm3
, of CO2 produced at room temperature and
pressure.
volume = ............................................... cm3
[1]
(iii) Calculate the volume of 2.50 mol dm–3 HNO3 needed to neutralise 2.68 g of
CaCO3.
volume = ............................................... cm3
[2]
[Total 5 marks]
i have done both first ones however i got confused on iii) please can someone explain to me what i should do
Calcium carbonate, CaCO3, reacts with acids such as nitric acid.
A student neutralised 2.68 g of CaCO3 with 2.50 mol dm–3 nitric acid, HNO3.
The equation for this reaction is shown below.
CaCO3(s) + 2HNO3(aq) → Ca(NO3)2(aq) + CO2(g) + H2O(l)
(i) Determine the amount, in mol, of CaCO3 reacted.
amount = ............................................... mol
(ii) Calculate the volume, in cm3
, of CO2 produced at room temperature and
pressure.
volume = ............................................... cm3
[1]
(iii) Calculate the volume of 2.50 mol dm–3 HNO3 needed to neutralise 2.68 g of
CaCO3.
volume = ............................................... cm3
[2]
[Total 5 marks]
i have done both first ones however i got confused on iii) please can someone explain to me what i should do
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Fredly
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Hi Mirams,
i) 0.0268 mol
ii) 643 cm3
iii) 21.4 cm3
Those are the answers I got.
Basically after calculating the mol of Calcium Carbonate (which is the answer to i) you have a 1:2 molar ratio in the equation for Ca Carbonate to Nitric Acid. So we do 0.0268*2 = 0.0536 which is the mol of nitric acid.
Just some advice always pay attention to units with these equations because you can be easily thrown off.
We now can use an equation that involves both concentration volume and mol to work out the conc, which would be the n/c = v
We are asked the calculate the volume of HNO3 required to neutralise CaCO3. We are told the conc, which is 2.5 moldm3 and we've worked out the mol which is 0.0536mol. So you just do 0.0536mol/2.5moldm3. You would get 0.02144dm3, remember mol units cancel in the division. The answer wants units in cm3 so you *1000. 0.2144* 1000 = 21.44cm3. Just go to 3 sig figs so 21.4.
I hope this has helped. Any question let me know.
i) 0.0268 mol
ii) 643 cm3
iii) 21.4 cm3
Those are the answers I got.
Basically after calculating the mol of Calcium Carbonate (which is the answer to i) you have a 1:2 molar ratio in the equation for Ca Carbonate to Nitric Acid. So we do 0.0268*2 = 0.0536 which is the mol of nitric acid.
Just some advice always pay attention to units with these equations because you can be easily thrown off.
We now can use an equation that involves both concentration volume and mol to work out the conc, which would be the n/c = v
We are asked the calculate the volume of HNO3 required to neutralise CaCO3. We are told the conc, which is 2.5 moldm3 and we've worked out the mol which is 0.0536mol. So you just do 0.0536mol/2.5moldm3. You would get 0.02144dm3, remember mol units cancel in the division. The answer wants units in cm3 so you *1000. 0.2144* 1000 = 21.44cm3. Just go to 3 sig figs so 21.4.
I hope this has helped. Any question let me know.
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