Answer:
0.4076 g
Explanation:
Kp is the equilibrium constant based on pressure and depends only on gas substances. For a generic reaction
aA + bB ⇄ cC + dD
[tex]Kp = \frac{(pC)^c*(pD)^d}{(pA)^a*(pB)^b}[/tex], where pX is the pressure of X in equilibrium.
For the reaction Kp = pCO₂
pCO₂ = 0.026 atm
The system is in equilibrium at the beginning. The compression occurs at a constant temperature, so using Boyle's law
P1V1 = P2V2
0.026*10 = P2*0.1
P2 = 2.6 atm
The reaction will reach again the equilibrium, and pCO₂ = 0.026 atm, then the rest will form MgCO₃, which will be 2.6 - 0.026 = 2.574 atm.
By the ideal gas law:
PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant (0.082 atm*L/mol*K), and T is the temperature.
2.574*0.1 = n*0.082*650
53.3n = 0.2574
n = 4.83x10⁻³ mol
The stoichiometry of the reaction is 1 mol of MgCO₃ for 1 mol of CO₂, so it will form 4.83x10⁻³ mol of MgCO₃ .
The molar mass is:
MgCO₃: 24 g/mol of Mg + 12 g/mol of C + 3*16 g/mol of O = 84 g/mol
The mass formed is the molar mass multiplied by the number of moles:
m = 84x4.83x10⁻³
m = 0.4076 g
The mass of [tex]\rm MgCO_3[/tex] (magnesium carbonate) that is formed is 0.4076 g.
Magnesium carbonate is an inorganic salt that white and solid in structure.
The equation is
aA + bB ⇄ cC + dD
Step1: By the Boyle's law
P1V1 = P2V2
[tex]\rm 0.026\times 10 = P_2\times0.1[/tex]
P2 = 2.6 atm.
For MgCO₃, which will be 2.6 - 0.026 = 2.574 atm
Step2: By the ideal gas law:
PV = nRT
[tex]2.574\times0.1 = n\times0.082\times650\\\\53.3n = 0.2574\\\\n = 4.83\times10^-^3 mol[/tex]
Step3: calculating the molar mass
[tex]\rm MgCO_3 = 24 g/mol \;of \;Mg + 12 g/mol \;of\; C + 3\times16 g/mol\; of\; O = 84 g/mol[/tex]
Step4: Calculating the mass
[tex]m = 84\times4.83x10^-^3\\\\m = 0.4076 g[/tex]
Thus, the mass is 0.4076 g.
Learn more about Magnesium carbonate, here
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