what volume of carbon dioxide will be collected at stp when 25.6 g of calcium carbohydrate undergoes decomposition? Balanced reaction: CaCO2 -> CaO + CO2

V = 5.74 L

Explanation:

Given data:

Volume of carbon dioxide produced = ?

Temperature and pressure = standard

Mass of calcium carbonate = 25.6 g

Solution:

Chemical equation:

CaCO₃      →        CaO + CO₂

Number of moles of calcium carbonate:

Number of moles = mass / molar mass

Molar mass of calcium carbonate = 100 g/mol

Number of moles = 25.6 g/ 100 g/mol

Number of moles = 0.256 mol

Now we will compare the mole of calcium carbonate with carbon dioxide.

                       CaCO₃           :            CO₂

                            1                :             1

                         0.256          :         0.256

Volume of  CO₂:

PV = nRT

R = general gas constant = 0.0821 atm/L / mol.K

By putting values,

1 atm× V = 0.256 mol × 0.0821 atm/L / mol.K× 273.15 K

V = 5.74 atm/L / 1 atm

V = 5.74 L

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml