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Question

The equation for the reaction of iron and oxygen gas to form rust (Fe2O3)\left(\mathrm{Fe}_2 \mathrm{O}_3\right) is written as

4Fe(s)+3O2(g)2Fe2O3(s)ΔH=1.7×103 kJ4 \mathrm{Fe}(s)+3 \mathrm{O}_2(g) \longrightarrow 2 \mathrm{Fe}_2 \mathrm{O}_3(s) \quad \Delta H=-1.7 \times 10^3 \mathrm{~kJ}

How many grams of rust form when 150 kJ are released?

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n(Fe2O3)ΔH2=150kJn(Fe_{2}O_{3})\cdot \dfrac{\Delta H}{2}=150kJ n(Fe2O3)=0.18molen(Fe_{2}O_{3})=0.18mole m=nMr=0.18mole159.7gmole1=28gm=n\cdot Mr=0.18mole\cdot 159.7gmole^{-1}=28g

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