Final answer:
To find the mass of oxygen in 25.0 g of KMnO4, calculate 40.5% of 25.0 g, which gives approximately 10.1 g. This is closest to option D) 10.6 g.
Step-by-step explanation:
To determine the mass of oxygen in 25.0 g of potassium permanganate (KMnO4), we first need to calculate the molar mass of KMnO4.
Potassium (K) = 39.10 g/mol
Manganese (Mn) = 54.94 g/mol
Oxygen (O) = 16.00 g/mol (since there are 4 oxygen atoms in KMnO4, 16.00 g/mol × 4 = 64.00 g/mol)
Total molar mass of KMnO4 = 39.10 g/mol (K) + 54.94 g/mol (Mn) + 64.00 g/mol (O) = 158.04 g/mol
The proportion of oxygen's mass in KMnO4 is 64.00 g/mol divided by the total molar mass of KMnO4, which is 64.00 g/158.04 g = 0.405 (or 40.5%).
Therefore, the mass of oxygen in 25.0 g of KMnO4 is 40.5% of 25.0 g, which is:
25.0 g × 0.405 = 10.125 g, which we can round to 10.1 g. This value is closest to option D) 10.6 g, but since option D is not an exact match, it is important to consider that the closest match may depend on rounding or significant figures used in the calculation.