For the reaction below, = 1.16 at 800.°c. caco₃(s) cao(s) + co₂(g) if a 25.0-g sample of caco₃ is put into a 14.4 l container and heated to 800°c, what percentage by mass of the caco₃ will react to reach equilibrium?
(a) 34.1%
(b) 100.0%
(c) 14.6%
(d) 65.9%

76%

Explanation:

For a reaction, Kp is the equilibrium constant based on the pressure, and it depends on only the partial pressure of the gas substances. It is the multiplication of the partial pressure of the products elevated by their coefficients, divided by the multiplication of the partial pressure of the reactants elevated by their coefficients.

For the given reaction

CaCO₃(s) ⇄ CaO(s) + CO₂(g)

Kp = pCO₂

pCO₂ = 1.16 atm

By the law of ideal gas:

P*V = n*R*T

Where P is the pressure, V is the volume, n is the number of moles, R is the gas constat (0.082 atm.L/mol.K), and T is the temperature (800ºC + 273 = 1073 K)

1.16*14.4 = n*0.082*1073

87.986n = 16.704

n = 0.1898 mol of CO₂.

Because the stoichiometry of the reaction is 1:1:1, the number of moles of CaCO₃ that reacted is 0.1898 mol. The molar mass of CaCO₃ is:

40 g/mol of Ca + 12 g/mol of C + 3*16 g/mol of O = 100 g/mol

mass = mol*molar mass

mass = 0.1898*100

mass = 18.98 g

The percentage by mass that will react is:

(mass that reacts/ initial mass) x 100%

(18.98/25)x100%

76%

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it