What mass of manganese can be obtained by reducing 150 g of manganese oxide

What mass of manganese can be obtained by reducing 150 g of manganese oxide 4 with carbon monoxide 2, if the mass fraction of the yield is 90%?

m (MnO3) = 150 g.

Let’s determine the mass of the manganese obtained. We write down the solution.

First, we write down the equation of reactions.

MnO3 + 3CO = Mn + 3CO2.

We find by the reaction equation the mass of the manganese obtained. We write down the solution.

First calculate the molar mass of manganese oxide and manganese oxide.

M (MnO3) = 55 + 3 * 16 = 103 g / mol.

M (Mn) = 55 g / mol.

Next, we find the mass of manganese.

150 g MnO3 – x g Mn.

103 g / mol MnO3 – 55 g / mol Mn.

X = 150 * 55: 103 = 80 g.

80 g is 100%.

X g is 90%.

X = 80 * 90: 100 = 72 g.

Answer: the mass of manganese is 72 g.