Question

What is the maximum mass of aluminum oxide that could be produced along with 64.39 g iron?

Answer 1

The maximum mass of aluminum oxide produced along with 64.39 g of iron is 58.78 g, calculated by using stoichiometry and the molar masses of iron (Fe) and aluminum oxide (Al2O3) from the balanced thermite reaction.

Step-by-step explanation:

Calculating the Mass of Aluminum Oxide Produced

To calculate the maximum mass of aluminum oxide that can be produced along with 64.39 g of iron (Fe), we need to consider the stoichiometry of the thermite reaction, which involves aluminum (Al) reacting with iron(III) oxide (Fe2O3). The reaction can be represented as:

2Al + Fe2O3 → Al2O3 + 2Fe

From the balanced equation, we can see that 1 mole of Fe2O3 produces 2 moles of Fe. Using the molar mass of Fe (55.85 g/mol), we can convert the given mass of iron (64.39 g) to moles:

Moles of Fe = 64.39 g Fe × (1 mol Fe / 55.85 g Fe) = 1.153 moles of Fe

According to the stoichiometry of the reaction, 1.153 moles of iron comes from 0.5765 moles of Fe2O3 (as every 2 moles of Fe come from 1 mole of Fe2O3). Now, we can find how much Al2O3 can be produced from 0.5765 moles of Fe2O3, as the molar ratio of Fe2O3 to Al2O3 is 1:1:

Moles of Al2O3 = 0.5765 moles of Fe2O3 (1 mole of Al2O3 / 1 mole of Fe2O3) = 0.5765 moles of Al2O3

Using the molar mass of Al2O3 (101.96 g/mol), we can convert the moles of Al2O3 to mass:

Mass of Al2O3 = 0.5765 moles Al2O3 × (101.96 g Al2O3 / 1 mol Al2O3) = 58.78 g of Al2O3 (to three significant figures)

Therefore, the maximum mass of aluminum oxide produced along with 64.39 g of iron is 58.78 g.

Answer 2

58.78 g Al₂O₃

Step-by-step explanation:

To find the mass of aluminum oxide (Al₂O₃) produced, you need to (1) convert grams Fe to moles Fe (via the molar mass), then (2) convert moles Fe to moles Al₂O₃ (via the mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via the molar mass). It is important that the conversions/ratios are arranged in a way that allows for the cancellation of units. The final answer should have 4 sig figs like the given value (64.39 g).

Molar Mass (Fe): 55.845 g/mol

Molar Mass (Al₂O₃): 2(26.982 g/mol) + 3(15.998 g/mol)

Molar Mass (Al₂O₃): 101.958 g/mol

Fe₂O₃(s) + 2 Al(s) -----> 2 Fe(l) + 1 Al₂O₃(s)
^ ^

64.39 g Fe 1 mole 1 mole Al₂O₃ 101.958 g
------------------- x ------------------ x ---------------------- x ------------------ =
55.845 g 2 moles Fe 1 mole

= 58.78 g Al₂O₃