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What is the standard enthalpy of formation of MgO(s) if 300.9 kJ is evolved when 20.15 g of MgO(s) is formed by the combustion of magnesium under standard conditions?

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$- 597.3 kJ {mol}^{- 1}$

$- 300.9 kJ {mol}^{- 1}$

$+ 300.9 {kJmol}^{- 1}$

$+ 597.3 kJ {mol}^{- 1}$

answer is A.

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Detailed Solution

$Mg (s) + 1 / 2 O_{2} (ρ) ⟶ {Mg}_{} O (s) {ΔH}_{r} = Δ_{f} i^{+} (MgO)$

$\begin{array}{l} \underset{{MgO}_{}}{W} = 20 .15 ρ \\ n_{mgO} = \frac{20 .15}{40} moles \end{array}$

$\begin{array}{l} {ΔH}_{r} = \frac{20 .15}{40} \times {ΔH}^{°}_{f (MgO)} \\ - 300 .19 \times \frac{40}{20 .15} = {ΔH}_{f (MgO)}^{\circ} \end{array}$

${ΔH}_{f}^{0} (MgO) = - 596 kJ$

$- \frac{300.9}{20.15} \times 40 = - 597.3 kJ / mol$

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