What is the theoretical yield in grams of MgO if 2.54 g Mgmetal reacts with exce

Question

What is the theoretical yield in grams of MgO if 2.54 g Mgmetal reacts with excess O2 ? What is the theoretical yield of Mg3N2if the same amount of Mg reacts with excess N2? What is the theoretical yield in grams of MgO if 2.54 g Mgmetal reacts with excess O2 ? What is the theoretical yield of Mg3N2if the same amount of Mg reacts with excess N2?

Explanation / Answer

2Mg + O2 -> 2MgO n(Mg) = m/M = 2.54/24.31 = 0.1045mol n(MgO) = n(Mg) = 0.1045mol n(MgO) = m/M m(MgO) = nM = 0.1045 x (24.31+16) = 4.21g 3Mg + N2 -> Mg3N2 n(Mg) = m/M = 2.54/24.31 = 0.1045mol n(Mg3N2) = 1/3 x n(Mg) = 0.03483mol n(Mg3N2) =m/M m(Mg3N2) = nM = 0.03483 x (24.31 x 3 + 14.01x 2) = 3.52g Hope this helps!

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