Question

For the reaction caco3 kp is 1.16 at 800°c if 20 g of caco3 were keep in 10 lit container and heating what

Answer 1

Given vaue of Kp =1.16 atm

Kp = pCO₂

=partial pressure of CO₂

=1.16 atm

Number of moles of CO2 released can be calculated using ideal gas equation,
Ideal gas equation will be

$PV=nRT$

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

So putting all in the equation:

P=1.16 atm

V=10 L

R=0.0821 atm L mol⁻¹ K
⁻¹

T=800 C= 1073 K

$n = \frac{PV}{RT}$

$= \frac{1.16 \times 10)} {0.082 \times 1073}$

= 0.13 mol of CO₂

It can be seen from the equation that,

CaCO₃----->CaO+CO₂

So,, one mole of CaCO₃ will form one mole of CO₂

So, 0.13 mol of CaCO₃ has reacted.

Mass of CaCO₃ reacted = 0.13 x 100 g = 13 g

Therefore, percentage of CaCO₃ reacted =$\frac{13}{20} \times 100$

= 65%

Answer 2

Answer:Since CaCO3 and CaO are solids, the value of Kp = partial pressure of CO2 = 1.16 atm.

Using ideal gas equation PV= nRT to calculate the number of moles of CO2 released.

n = PV/RT = 1.16 x 5.6) / 0.082 x 1073K

................= 0.0738 mol of CO2

From the equation, one mole of CaCO3 will form one mole of CO2

Hence, 0.0738 mol of CaCO3 has reacted.

Mass of CaCO3 reacted = 0.0738 x 100 g = 7.38 g

Terefore, percentage of CaCO3 reacted = 7.38/48 x 100%

.........................................................= 15.38%

Since CaCO3 and CaO are solids, the value of Kp = partial pressure of CO2 = 1.16 atm.

Using ideal gas equation PV= nRT to calculate the number of moles of CO2 released.

n = PV/RT = 1.16 x 5.6) / 0.082 x 1073K

................= 0.0738 mol of CO2

From the equation, one mole of CaCO3 will form one mole of CO2

Hence, 0.0738 mol of CaCO3 has reacted.

Mass of CaCO3 reacted = 0.0738 x 100 g = 7.38 g

Terefore, percentage of CaCO3 reacted = 7.38/48 x 100%

.........................................................= 15.38%

Explanation: