Final answer:
To determine the mass of Mn₂O₃ produced, we need to use the balanced chemical equation and calculate the number of moles of MnO₂ in 46.8 g. Then, using the stoichiometry from the equation, we can calculate the moles of Mn₂O₃ produced and convert it to grams.
Step-by-step explanation:
To determine the mass of Mn₂O₃ produced from the complete reaction of 46.8 g of MnO₂, we need to use the balanced chemical equation:
Zn + 2MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃
We can see that for every 2 moles of MnO₂ in the reaction, we get 1 mole of Mn₂O₃. The molar mass of MnO₂ is 86.936 g/mol, so we can calculate the moles of MnO₂ in 46.8 g:
moles of MnO₂ = mass of MnO₂ / molar mass of MnO₂ = 46.8 g / 86.936 g/mol = 0.537 moles
Since the ratio is 2:1, the moles of Mn₂O₃ produced will be half that of MnO₂:
moles of Mn₂O₃ = 0.537 moles / 2 = 0.2685 moles
Finally, we can calculate the mass of Mn₂O₃:
mass of Mn₂O₃ = moles of Mn₂O₃ * molar mass of Mn₂O₃ = 0.2685 moles * 157.87 g/mol = 42.41 g