Answer:
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