I guess the answer is MgO4S
Answer:
The answer to your question is ΔH = -964.2 kJ
Explanation:
Data
Reaction 1 X + 12O₂ ⇒ XO ΔH = -607.3 kJ
Reaction 2 XCO₃ ⇒ XO + CO₂ ΔH = 356.9 kJ
ΔH for Reaction 3 X + 12O₂ + CO₂ ⇒ XCO₃
Process
1.- Write the Reaction 1 the same
2.- Invert the order of reaction 2, and change the sign of the Enthalpy
Reaction 1 X + 12O₂ ⇒ XO ΔH = -607.3 kJ
Reaction 2 XO + CO₂ ⇒ XCO₃ ΔH = - 356.9 kJ
3.- Sum up the reactions
X + 12O₂ + XO + CO₂ ⇒ XO + XCO₃
4.- Simplify
X + 12O₂ + CO₂ ⇒ XCO₃ ΔH = -964.2 kJ
<span> (0.20948 x 31.998) / (
(0.78084 x 28.013) +
(0.20948 x 31.998) +
(0.00934 x 39.948) +
(0.000375 x 44.0099) +
(0.00001818 x 20.183) +
(0.00000524 x 4.003) +
(0.000002 x 16.043) +
(0.00000114 x 83.80) +
(0.0000005 x 2.0159) +
(0.0000005 x 44.0128) ) = 0.23140 = 23.140% O by mass </span>
