Question

How many grams Mn2O3 would be produced from the complete reaction of 46.8 g of MnO2 ?

Zn + 2MnO2 + H2O Zn(OH) 2 +Mn2O3
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Answer 1

There should be at-least into this the code to get it out and get the back up and then go back into it and get a new one therefore it will be 53.5 grams of MN203

Answer 2

The balanced equation for the reaction is:

Zn + 2MnO2 + H2O → Zn(OH)2 + Mn2O3

Explanation:

The molar mass of MnO2 is 86.94 g/mol, so 46.8 g of MnO2 is equivalent to 46.8 / 86.94 = 0.536 moles of MnO2.

The molar mass of Mn2O3 is 157.88 g/mol, so 0.536 moles of Mn2O3 will be produced, which is equivalent to 0.536 * 157.88 = 85.3 g of Mn2O3.

Therefore, 85.3 g of Mn2O3 would be produced from the complete reaction of 46.8 g of MnO2.

Here is the calculation:

Mass of Mn2O3 produced = (Number of moles of Mn2O3 produced) * (Molar mass of Mn2O3)

= 0.536 moles * 157.88 g/mol

= 85.3 g