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.
Where should I start looking for actually?
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.
Thermochemistry Essay Rough Draft
To answer this question, we need to go step by step.
1. First, we need to calculate the number of moles of HCl used for neutralization. We know the concentration of the HCl solution (1.0M), and the volume used (25.00mL).
Moles of HCl = Concentration (M) x Volume (L)
Convert the volume to liters: 25.00mL = 0.025L
Moles of HCl = 1.0M x 0.025L = 0.025 moles of HCl
2. Next, we need to calculate the number of moles of NaOH used for neutralization. We know the concentration of the NaOH solution (0.1020M), and the volume used (21.02mL).
Convert the volume to liters: 21.02mL = 0.02102L
Moles of NaOH = 0.1020M x 0.02102L = 0.002143 moles of NaOH
3. Due to the balanced chemical equation between HCl and NaOH, we know that 1 mole of HCl reacts with 1 mole of NaOH. Therefore, the moles of HCl used will be equal to the moles of NaOH used:
Moles of HCl = Moles of NaOH = 0.025 moles of HCl
4. Now, we can use the stoichiometry of the balanced chemical equation to find the moles of the metallic oxide (MO) that reacted with 0.025 moles of HCl.
From the balanced equation, we know that 1 mole of MO reacts with 2 moles of HCl.
Moles of MO = (Moles of HCl x Molecular weight of HCl) / 2
The molecular weight of HCl = 1.007 + 35.453 = 36.46 g/mol. (Add the atomic weights of hydrogen and chlorine).
Moles of MO = (0.025 x 36.46) / 2
5. Use the given mass (1.600g) of the metallic oxide to find the mass of the oxide that combines with 1 mole of HCl.
Mass of the oxide combining with 1 mole of HCl = (Mass of MO x Moles of MO) / 1
Mass of the oxide combining with 1 mole of HCl = (1.600 x 0.025 x 36.46) / 2
6. Finally, determine the formula mass of the oxide by dividing the mass of the oxide by the number of moles of the oxide that combines with 1 mole of HCl.
Formula mass of the oxide = Mass of the oxide / Moles of the oxide
Formula mass of the oxide = (1.600 x 0.025 x 36.46) / ((1.600 x 0.025 x 36.46) / 2)
7. The atomic mass of the metal can be determined using the formula mass of the oxide and the chemical formula of the oxide, which is not given in the question. You need additional information about the oxide's chemical formula or any other clue to perform this step.
So, to actually answer the question and determine the formula mass of the oxide and the atomic mass of the metal, you will need additional information about the chemical formula of the oxide or any other relevant details.