1g of mixture of calcium carbonate and magnesium carbonate gave 240mL of carbon dioxide at ntp. Calculate monoxide to carbon dioxide.
Answers
Answer:
Let the mass of.. CaC O 3
CaCO3
in the mixture = x g
Mass of
MgC
O
3
MgCO3
in the mixture
=(1−x)g
=(1-x)g
Step I. Calculation of volume of
C
O
2
CO2
evolved from
CaC
O
3
CaCO3
CaC
O
3
100 g
(s)
−
→
heat
CaO(s)+
C
O
2
(g)
22400 mL
CaCO3100 g(s)→heatCaO(s)+CO2(g)22400 mL
100 g of
CaC
O
3
CaCO3
evolve
C
O
2
(g)=22400mL
CO2(g)=22400mL
xg
xg
of
CaC
O
3
CaCO3
evolve
C
O
2
(g)=
(xg)
(100g)
×(22400mL)=(224x)mL
CO2(g)=(xg)(100g)×(22400mL)=(224x)mL
Step II. Calculation of volume of
C
O
2
CO2
evolved from
MgC
O
3
MgCO3
MgC
O
3
(s)
84 g
→MgO(s)+
C
O
2
(g)
22400 mL
MgCO3(s)84 g→MgO(s)+CO2(g)22400 mL
84 g of
MgC
O
3
MgCO3
evolve
C
O
2
(g)=22400mL
CO2(g)=22400mL
(1−x)g
(1-x)g
of
MgC
O
3
MgCO3
evolve
C
O
2
(g)=
(1−x)g
(84g)
×22400mL=266.67(1−x)mL
CO2(g)=(1-x)g(84g)×22400mL=266.67(1-x)mL
Step III. Calculation of percentage composition
The volume of
C
O
2
CO2
evolved
=(22x)mL+266.67(1−x)mL
=(22x)mL+266.67(1-x)mL
But volume of
C
O
2
CO2
actually evolved = 240 mL
On equating,we get
224x+266.67(1−x)=240
224x+266.67(1-x)=240
,
224x+266.67−266.67x=240or−42.67x=−26.67
224x+266.67-266.67x=240or-42.67x=-26.67
or
x=
26.67
42.67
=0.6250
x=26.6742.67=0.6250
∴
∴
Percentage of
CaC
O
3
CaCO3
in the mixture
=0.6250×100=62.50%
=0.6250×100=62.50%
Percentage of
MgC
O
3
MgCO3
in the mixture
=100−62.50=37.50%
=100-62.50=37.50%
.
Explanation: