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.
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