Final answer:
To find the mass of Mn₂O₃ produced from the reaction of 46.8 g of MnO₂, we use stoichiometry to first calculate moles of MnO₂, then use the mole ratio from the balanced equation, and finally convert the moles of Mn₂O₃ to grams, resulting in 42.5 g of Mn₂O₃.
Step-by-step explanation:
The student is asking how many grams of manganese(III) oxide (Mn₂O₃) would be produced from the complete reaction of 46.8 grams of manganese(IV) oxide (MnO₂) using the given chemical reaction:
Zn + 2MnO₂ + H₂O → Zn(OH)₂ + Mn₂O₃.
To answer this, we will perform stoichiometry.
First, we calculate the moles of MnO₂ reacted using its molar mass (86.94 g/mol).
Then, using the stoichiometry from the balanced equation (2 moles of MnO₂ produce 1 mole of Mn₂O₃), we determine the moles of Mn₂O₃ formed.
Finally, we convert the moles of Mn₂O₃ into grams using its molar mass (157.88 g/mol).
To calculate the mass of Mn₂O₃:
- Calculate moles of MnO₂: 46.8 g MnO₂ × (1 mol MnO₂ / 86.94 g MnO₂) = 0.538 mol MnO₂
- Using stoichiometry, calculate moles of Mn₂O₃: 0.538 mol MnO₂ × (1 mol Mn₂O₃ / 2 mol MnO₂) = 0.269 mol Mn₂O₃
- Convert moles of Mn₂O₃ to grams: 0.269 mol Mn₂O₃ × (157.88 g Mn₂O₃ / 1 mol Mn₂O₃) = 42.5 g Mn₂O₃
Therefore, 42.5 grams of Mn₂O₃ would be produced from the complete reaction of 46.8 grams of MnO₂.