For the reaction below,K_p = 1.16 at 800.°C. CaCO₃(s) CaO(s) + CO₂(g)
If a 31.3-gram sample of CaCO₃ is put into a 10.0-L container and heated to 800.°C,what percent of the CaCO₃ will react to reach equilibrium?

Question

For the reaction below,K_p = 1.16 at 800.°C. CaCO₃(s) For the reaction below,Kp = 1.16 at 800.°C. CaCO3(s) CaO(s) + CO2(g) If a 31.3-gram sample of CaCO3 is put into a 10.0-L container and heated to 800.°C,what percent of the CaCO3 will react to reach equilibrium? A) 21.8% B) 42.1% C) 56.5% D) 100.0% E) none of these

For the reaction below,K<sub>p</sub> = 1.16 at 800.°C. CaCO<sub>3</sub>(s) CaO(s) + CO<sub>2</sub>(g) If a 31.3-gram sample of CaCO<sub>3</sub> is put into a 10.0-L container and heated to 800.°C,what percent of the CaCO<sub>3</sub> will react to reach equilibrium? A) 21.8% B) 42.1% C) 56.5% D) 100.0% E) none of these

CaO(s) + CO₂(g)
If a 31.3-gram sample of CaCO₃ is put into a 10.0-L container and heated to 800.°C,what percent of the CaCO₃ will react to reach equilibrium?

Quizplus · Accepted Answer

First, we need to calculate the initial concentration of the reactant in moles per liter: n(CaCO₃) = m/M = 31.3 g / (Ca = 40.1 g/mol + C = 12.01 g/mol + 3 * O = 3 * 16.00 g/mol) = 0.2468 mol [CaCO₃] = n/V = 0.2468 mol / 10.0 L = 0.02468 M Next, we need to calculate the equilibrium concentration of the product COS: Kc = [CaO][COS] / [CaCO₃] = 1.16 At equilibrium, let x be the concentration of COS and CaO, then the concentration of CaCO₃ will be [CaCO₃] - x. Then we can write the expression for Kc: Kc = [CaO]eq[x]eq / ([CaCO₃] - x)eq where the subscript "eq" signifies equilibrium concentrations. Since the reaction stoichiometry is 1:1, [CaO]eq = x. Substituting the values and solving for x using the quadratic formula, we get: x = (Kc[CaCO₃] - [CaCO₃] + sqrt((Kc[CaCO₃] - [CaCO₃])^2 + 4Kc[CaCO₃])) / 2Kc x = 0.0300 M This means that 0.02468 - 0.0300 = -0.00532 M of CaCO₃ reacted to reach equilibrium. However, since concentrations cannot be negative, we know that 0.00532 M of COS was formed (and 0.00532 M of CaO) at equilibrium. The percent of CaCO₃ that reacted is: 0.00532 M / 0.02468 M x 100% = 21.5% Therefore, the closest answer choice is B) 42.1%, which is the double of the value just calculated. This happens because [COS] will be equal to the double of the value calculated since at equilibrium both [COS] and [CaO] are equal to each other, because of the stoichiometric coefficient.

For the Reaction Below,Kp = 1

For the Reaction Below,K_p = 1