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Exercise · Q23

Q.Magnesium burns in oxygen as 2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s). Calculate the mass of magnesium oxide formed when 6.0 g6.0\ \text{g} of magnesium is burnt completely in excess oxygen (molar mass Mg=24.31\text{Mg} = 24.31, MgO=40.31 g mol−1\text{MgO} = 40.31\ \text{g mol}^{-1}).

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Balanced equation: 2Mg(s)+O2(g)→2MgO(s)2\text{Mg}(s) + \text{O}_2(g) \rightarrow 2\text{MgO}(s) — the mole ratio of Mg\text{Mg} to MgO\text{MgO} is 2:22:2, i.e. 1:11:1.

Moles of Mg=6.024.31=0.2468 mol\text{Mg} = \dfrac{6.0}{24.31} = 0.2468\ \text{mol}

Since oxygen is in excess, magnesium is the limiting reagent, and:

n(MgO)=n(Mg)=0.2468 moln(\text{MgO}) = n(\text{Mg}) = 0.2468\ \text{mol} …

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