Final answer:
1.55 g of manganese(IV) oxide will react to produce 1.91 g of aluminum oxide.
Step-by-step explanation:
To solve this problem, we need to use the balanced chemical equation and mole ratios. The balanced equation is:
3MnO2(s) + 4Al(s) → 2Al2O3(s) + 3Mn(s)
From the equation, we can see that 3 moles of MnO2 react with 4 moles of Al to produce 2 moles of Al2O3. We are given 1.55 g of MnO2, so we first need to convert it to moles using its molar mass:
Molar mass of MnO2: 54.938 g/mol
1.55 g MnO2 × (1 mol MnO2 / 54.938 g MnO2) = 0.0282 mol MnO2
Now, we can use the mole ratio to find the moles of Al2O3:
Mole ratio of MnO2 to Al2O3: 3:2
0.0282 mol MnO2 × (2 mol Al2O3 / 3 mol MnO2) = 0.0188 mol Al2O3
Finally, we can convert the moles of Al2O3 to grams using its molar mass:
Molar mass of Al2O3: 101.96 g/mol
0.0188 mol Al2O3 × (101.96 g Al2O3 / 1 mol Al2O3) = 1.91 g Al2O3
Therefore, 1.91 grams of aluminum oxide will be produced.