Question

determining and interpreting the enthalpy of the magnesium combustion reaction
earlier in the lesson, you determined that the enthalpy for magnesium combustion could be computed using hesss law. the formula derived from the reaction equations was: $delta h=delta h_1 - delta h_2+delta h_3$
the enthalpy of the magnesium combustion is kj/mol.
magnesium combustion
reaction equation $delta h$ (kj/mol)
reaction 1: $mg(s)+2hcl(aq)
ightarrow mgcl_2(aq)+h_2(g)$ -450
reaction 2: $mgo(s)+2hcl(aq)
ightarrow mgcl_2(aq)+h_2o(l)$ -81
reaction 3: $h_2(g)+\frac{1}{2}o_2(g)
ightarrow h_2o(l)$ -200
magnesium combustion: $mg(s)+\frac{1}{2}o_2(g)
ightarrow mgo(s)$ ?

Explanation:

Step1: Identify the given $\Delta H$ values

$\Delta H_1=-450$ kJ/mol, $\Delta H_2 = - 81$ kJ/mol, $\Delta H_3=-200$ kJ/mol

Step2: Apply the formula $\Delta H=\Delta H_1-\Delta H_2+\Delta H_3$

$\Delta H=-450-(-81)+(-200)$
$=-450 + 81-200$
$=-569$ kJ/mol

Answer:

There seems to be an error in the provided answer - options as the calculated value is - 569 kJ/mol and it is not among the given options (-817, -654, -245). If we assume there is some other correct - way of calculation based on the context not fully shown here, we need more information. But based on the formula $\Delta H=\Delta H_1-\Delta H_2+\Delta H_3$ with the given values, the result is - 569 kJ/mol.