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Urgent about reacting mass !!! (Chem)

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When 22.9g of an oxide of metal M is completely reduced by hydrogen, 7.73g of water is obtained. What is the empirical formular of the oxide ? [Relative atomic mass of M = 55.8] A. MO B. M2O C. M2O3 D. M3O2

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Let the formula of metal oxide be MXOY. MXOY(s) + Y H2(g) → X M(s) + Y H2O(l) Molar mass of H2O = 2 + 16 = 18 no. of moles of H2O = 7.73/18 = 0.429 Molar mass of MXOY = 55.8X + 16Y no. of moles of MXOY = 22.9/(55.8X + 16Y) mole ratio H2O : MXOY = Y : 1 0.429 : 22.9/(55.8X + 16Y) = Y : 1 0.429 (55.8X + 16Y) : 22.9 = Y : 1 0.429 (55.8X + 16Y) = 22.9Y 55.8X = 37.38Y X : Y = 37.38 : 55.8 X : Y = 37.38/37.38 : 55.8/37.38 X : Y = 1 : 1.49 X : Y = 2 : 3 ∴ The empirical formula of the oxide is M2O3 ∴ The answer is C. M2O3 希望幫到您！

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