Final answer:
The mass of aluminum oxide produced from manganese(IV) oxide, a balanced chemical reaction is necessary .The mass of aluminum oxide (Al2O3) produced from 1.74 g of manganese(IV) oxide (MnO2) is 2.04 g.
Step-by-step explanation:
To determine the mass of aluminum oxide (Al2O3) produced from 1.74 g of manganese(IV) oxide (MnO2), we need to use the balanced chemical equation and the molar ratios between the reactants and products.
The balanced equation is:
2 MnO2 + 4 Al → Mn2O3 + 2 Al2O3
From the equation, we can see that 2 moles of MnO2 react to form 2 moles of Al2O3. We can then use the molar mass of MnO2 and the molar mass of Al2O3 to calculate the mass of Al2O3.
First, calculate the number of moles of MnO2 using its molar mass:
Molar mass of MnO2 = 86.94 g/mol
Number of moles = mass / molar mass = 1.74 g / 86.94 g/mol = 0.02 mol
Since 2 moles of MnO2 react to form 2 moles of Al2O3, the number of moles of Al2O3 produced is also 0.02 mol.
Finally, calculate the mass of Al2O3 using its molar mass:
Molar mass of Al2O3 = 101.96 g/mol
Mass of Al2O3 = number of moles x molar mass = 0.02 mol x 101.96 g/mol = 2.04 g