1.80g of metal oxide required 833ml of hydrogen at NTP to be reduced to its metal find the equivalent weight of oxide and the metal

833 mL of H2 = 0.03719 mole of H2 M^+2 MO + H2 --> M + H2O M2O + H2 --> 2 M + H2O 1.8 g MO / 0.03719 mole = 48.4 g/mol of MO or M2O M = 32.4 g/mol M^+4 MO2 + 2 H2 --> M + 2 H2O 1.8 g MO2 / 0.01860 mole = 96.8 g/mol M = 64.8 g/mol M^+3 M2O3 + 3 H2 --> 2 M + 3 H2O 1.8 g M2O3 / 0.01240 = 145 M = 48.5 g/mol M^+1 M2O + H2 --> 2 M + H2O 1.8 g M2O / 0.03719 = 48.4 M = 24.2 g/mol Titanium(III)oxide Ti2O3 is fairly close. Titanium is 47.9 g/mol vs 48.5 calculated value