Pure calcium carbonate and dilute hydrochloric acid are reacted with 2 litres of carbon
dioxide was collected at 27ºC and normal pressure.
(i) The mass of salt required
Answers
Explanation:
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Answer:
The mass of salt required is, 8.12 grams
(b) The mass of the acid required is, 5.93 grams
Explanation : Given,
Volume of carbon dioxide gas = 2 L
Temperature of carbon dioxide gas = 27^oC=273+27=300K27
o
C=273+27=300K
Pressure of carbon dioxide gas = 1 atm
First we have to calculate the moles of carbon dioxide gas by using ideal gas equation.
PV=nRTPV=nRT
where,
P = pressure of carbon dioxide gas
V = volume of carbon dioxide gas
T = temperature of carbon dioxide gas
n = number of moles of carbon dioxide gas
R = gas constant = 0.0821 L.atm/mole.K
(1atm)\times (2L)=n\times (0.0821L.atm/mole.K)\times (300K)(1atm)×(2L)=n×(0.0821L.atm/mole.K)×(300K)
n=0.0812molen=0.0812mole
(a) Now we have to calculate the mass of salt required.
The balanced chemical reaction is,
CaCO_3(s)+2HCl(l)\rightarrow CaCl_2(s)+H_2o(l)+CO_2(g)CaCO
3
(s)+2HCl(l)→CaCl
2
(s)+H
2
o(l)+CO
2
(g)
From the balanced reaction we conclude that,
As, 1 mole of CO_2CO
2
obtained from 1 mole of CaCO_3CaCO
3
So, 0.0812 mole of CO_2CO
2
obtained from 0.0812 mole of CaCO_3CaCO
3
\text{Mass of }CaCO_3=\text{Moles of }CaCO_3\times \text{Molar mass of }CaCO_3=0.0812mole\times 100g/mole=8.12gMass of CaCO
3
=Moles of CaCO
3
×Molar mass of CaCO
3
=0.0812mole×100g/mole=8.12g
Therefore, the mass of the salt required is, 8.12 grams
(b) Now we have to calculate the mass of acid required.
The balanced chemical reaction is,
CaCO_3(s)+2HCl(l)\rightarrow CaCl_2(s)+H_2o(l)+CO_2(g)CaCO
3
(s)+2HCl(l)→CaCl
2
(s)+H
2
o(l)+CO
2
(g)
From the balanced reaction we conclude that,
As, 1 mole of CO_2CO
2
obtained from 2 mole of HClHCl
So, 0.0812 mole of CO_2CO
2
obtained from 0.1624 mole of HClHCl
\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl=0.1624mole\times 36.5g/mole=5.93gMass of HCl=Moles of HCl×Molar mass of HCl=0.1624mole×36.5g/mole=5.93g
Therefore, the mass of the acid required is, 5.93 grams
Explanation:
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