0.100 mol of CaCO3 and 0.100 mol CaO are placed in an 10.0 L evacuated container and heated to 385 K. When equilibrium is reached the pressure of CO2 is 0.220 atm. 0.230 atm of CO2 is added, while keeping the temperature constant and the system is allowed to reach again equilibrium. What will be the final mass of CaCO3

10.32g

Explanation:

First thing's first, lets bring out the balanced chemical equation representing the reaction.

CaCO3(s) ⇄ CaO(s) + CO2(g)

From the question;

reaction constant K = [CO2(g)] = 0.220 atm.

To obtain number of moles of CO2, using ideal gas fomula, we have;

n = PV/RT = 0.220*10 / (0.08206*385) = 0.0696(mol)

From the stoichiometry of the equation above;

This is the amount of CaCO3 which has been converted to CaO before additional 0.230 atm CO2(g) is added.

After the additional 0.230 atm CO2(g), the equilibrium CO2 pressure is still 0.220 atm.  Therefore all this additional CO2 would completely convert to CaCO3:

n = PV/RT = 0.230*10.0/(0.08206*385) = 0.0728 (mol)

Hence the total CaCO3 after equilibrium is reestablished is:

0.100 - 0.0696 + 0.0728 (mol) = 0.1032 mol

Mass = Number of moles * Molar mass

Mass = 0.1032 * 100 = 10.32g

i believe it is-

d.) gravity pulled particles of dust and gas together to form planets.

as given the equation is

2al + 6 hcl > 2 alcl3 + 3 h2(g)

as per equation two moles of aluminum reacts with six moles of hcl to give two moles of alcl3 and three moles of hydrogen gas

given: three moles of al are present and 7 moles of hcl

for three moles of aluminum nine moles of hcl will be required

however as 7 moles of hcl are present so the limiting reagent is hcl

hcl is going to decide the number of moles of product formed

when six moles of hcl are present = 3 moles of h2 are formed

when one mole of hcl is present == 3/6 mole sof h2 will be formed

when 7 moles of hcl are present = 3 x 7/ 6 moles of h2 will be formed = 3.5 moles of h2