Question

calculate the volume, mass and
number of molecules of carbon
dioxide is liberated when 8
grames of calcium carbonate is
docomposaed on heating.(atomic masses of ca =40u,c=12u and o=16u)​

Answer 1

Answer:

Chemical reaction involved is:

CaCO  

3

​  

→CaO+CO  

2

​  

 

Molar mass of CaCO  

3

​  

=40+12+16×3=100 g

According to the balanced reaction 100 g of CaCO  

3

​  

 produces 1 mol of CO  

2

​  

 i.e. 22.4 L at STP.

∴50 g CaCO  

3

​  

 will produce  

100

22.4

​  

×50=11.2 L CO  

2

​  

 at STP.

Explanation:

Answer 2

Good Noon

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Here's your answer

CaCO₃ ------> CaO + CO₂

In this reaction,

1 mole CaCO₃ decomposes to give 1 mole CO₂

Molecular mass of CaCO₃ = 40+12+48 = 100

Number of moles of CaCO₃ present = Given mass/Molecular mass

                                                               = 8/100

Hence,

8/100 mole CaCO₃ decompose to give 8/100 moles of CO₂

Number of moles of CO₂ = Given mass/Molar mass

Given mass = Number of moles * Molar mass

                    = 8/100 * 44 = 352/100 = 3.52

Number of moles of CO₂ = Volume/22.4

Volume = Number of moles * 22.4

              = 8/100 * 22.4 = 179.2/100 = 1.792 L

Number of moles of CO₂ = Number of particles/ Avogadro Number

Number of particles = Number of moles * 6.022 * 10²³

                                   = 8/100 * 6.022 * 10²³

                                   = 48.176 * 10²¹

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