MO + 2HCl ==> MCl2 + H2O
How many mols HCl did you start with? That's 0.1L x 1.00M = 0.1 mol.
How much was neutralized by NaOH? That's 0.02102L x 0.1020M = 0.02144 but that was for a 25.0 mL aliquot of thae original. So the ORIGINAL amount of HCl neutralized was 0.02144 x (500/25) = 0.04288. So how much HCl was used? That must be initial-used = 0.1 - 0.04288 = 0.05712 mol HCl used in the rxn with MO and since there were 2 mols HCl used for every 1 mol MO, mols MO must be 1/2 that or 0.05712/2 = 0.02856 mol MO in 1.6 g.
mols = grams/molar mass or
molar mass = grams/mols = 1.6/0.02856 = about 56 but you should do all of these calculations a little more carefully. So if the molar mass is about 56 and you know O is 16, that leaves what? about 50 for M. Hope this helps.
1.600g of a metallic oxide of type MO were dissolved in100mL 1.0M HCl. The resulting liquid was made up to 500mL with distilled water. 25.00mL of the solution ,then required 21.02mL of 0.1020 M of NaOH for neutralization. Calculate the mass of the oxide combining with 1 mole of HCl and hence determine the formula mass of the oxide and the atomic mass of the metal.
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