Final answer:
The percentage by mass of Mn3O4 in the ore is calculated to be 12.07%, which does not match any of the given options. There may be an error in the question or options provided.
Step-by-step explanation:
To calculate the percentage by mass of Mn3O4 in the ore, we first need to know the molar mass of Mn3O4, which is composed of 3 manganese (Mn) atoms and 4 oxygen (O) atoms. Using the periodic table, we calculate the molar mass as follows: 3(Mn) + 4(O) = 3(54.938) + 4(16.00) = 164.814 + 64.00 = 228.814 g/mol.
Since 200.0 kg (or 200,000 g) of manganese is obtained from 23 tonnes (or 23,000,000 g) of ore, we can presume that all this manganese comes from Mn3O4, and we can calculate the mass of Mn3O4 as:
Mass of Mn in Mn3O4 = 200,000 g (total mass of manganese obtained)
Molar mass of Mn = 54.938 g/mol
Molar mass of Mn3O4 = 228.814 g/mol
The molar ratio of Mn to Mn3O4 is 3:1, so we need to multiply the mass of Mn by the molar mass of Mn3O4 and then divide by the mass of 3 Mn to obtain the mass of Mn3O4.
Mass of Mn3O4 = (200,000 g × 228.814 g/mol) / (3 × 54.938 g/mol) = 2,775,547.2 g of Mn3O4
Now we can find the mass percentage of Mn3O4 in the ore:
Percentage by mass = (mass of Mn3O4 / mass of ore) × 100%
Percentage by mass = (2,775,547.2 g / 23,000,000 g) × 100% = 12.07%
None of the options given match the calculated value. There might be a mistake in the question or in the options provided.