Chemistry, asked by samreen1464, 11 months ago

1g of mixture of calcium carbonate and magnesium carbonate gave 240mL of carbon dioxide at ntp. Calculate monoxide to carbon dioxide.

Answers

Answered by harshitabharti
0

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:

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