Question -2Fe(s)+32O2(g)→Fe2O3(s)(△H=−193.4kJ)......(1) Mg(s)+12O2(g)→MgO(s)(△H=−140.2kJ)......(2) What is △∆H of the reaction? 3Mg+Fe2O3→3MgO+2Fe
Options:
A .  -272.3 kJ
B .  272.3 kJ
C .  272.2 kJ
D .  -227.2 kJ
Answer: Option D : D Equation (2) is multiplied by 3 2Fe(s)+32O2(g)→Fe2O3(s)△H=−193.4kJ......(1) 3Mg(s)+32O2(g)→3MgO(s)△H=−420.6kJ......(3) Subtracting equation (1) from (3) 3Mg(s)+Fe2O3→3MgO+2Fe△H=−420.6−(−193.4) = -227.2 kJ
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