At 900 oC, Kp = 1.04 for the reactionCaCO31(s) arrow CaO (s) + CO2 (g)At a low temperature, dry ice (solid CO2), calcium oxide, and calcium carbonate are introduced into a 50.0-L reaction chamber. The temperature is raised to 900 oC, resulting in the dry ice converting to gaseous CO2. For the following mixtures, will the initial amount of calcium oxide increase, decrease, or remain the same as the system moves toward equilibrium at 9000 oC?a. 655 gCaCO3, 95.0 gCaO, PCO2 = 2.55 atm. b. 780 gCaCO3, 1.00 gCaO, PCO2 = 1.o4 atm. c. 0.14 gCaCO3, 5000 gCaO, PCO2 = 1.04 atm. d. 715 gCaCO3, 813 gCaO, PCO2 = 0.211 atm.

a)the reaction will proceed to the left, the mass of CaO will decrease.

b)the reaction is in equilibrium, mass of CaO will not change.

c)the reaction is in equilibrium, mass of CaO will not change.

d) the reaction will proceed to the right, the mass of CaO will increase.

Explanation:

Step 1: Data given

At 900°C, Kp = 1.04  ( Kp is only for compounds that are in gaseous state)

Volume of the chamber = 50.0L

Temperature is raised to 900 °C

Step 2: The reaction

CaCO31(s) ⇔ CaO (s) + CO2 (g)

Kp for CO2 = 1.04

Step 3: First mix :  655g CaCO3, 95.0g CaO, PCO2 = 2.55 atm

We only consider CO2 here

pCO2 > Kp  this means the reaction will proceed to the left, the mass of CaO will decrease.

Step 4: 780g CaCO3, 1.00g CaO, PCO2 = 1.04 atm.

We only consider CO2 here

pCO2 = Kp  this means the reaction is in equilibrium, mass of CaO will not change.

Step 5: 0.14g CaCO3, 5000g CaO, PCO2 = 1.04 atm

We only consider CO2 here

pCO2 = Kp  this means the reaction is in equilibrium, mass of CaO will not change.

Step 6: 715g CaCO3, 813g CaO, PCO2 = 0.211 atm

We only consider CO2 here

pCO2 < Kp  this means the reaction will proceed to the right, the mass of CaO will increase.

so here we go

atomic mass

239 + 1 = y + 134 + 3 combine like terms

240 = y + 137             subtract 137 from both terms

240 - 137 = y

103 = y

atomic number

94 = z +54 + 0

94 = z + 54               subtract 54 from both sides.

40 = z

answer