Question

54.0g Al reacts with 64.0g O2 to form Al2O3 according to the equation.

4Al+3O2 = 2Al2O3
O2: 32 g/mol Al2O3: 102 g/mol
How many grams of Al2O3 form from 64.0 g O2?

[?]g Al2O3

Answer 1

136 g Al₂O₃

Step-by-step explanation:

Assuming you do not need to find the limiting reactant, to find the mass of Al₂O₃, you need to (1) convert grams O₂ to moles O₂ (via molar mass), then (2) convert moles O₂ to moles Al₂O₃ (via mole-to-mole ratio from equation coefficients), and then (3) convert moles Al₂O₃ to grams Al₂O₃ (via molar mass). It is important to arrange the conversions in a way that allows for the cancellation of units. The final answer should have 3 sig figs to match the sig figs of the given value (64.0 g).

Molar Mass (O₂): 32 g/mol

Molar Mass (Al₂O₃): 102 g/mol

4 Al + 3 O₂ -----> 2 Al₂O₃

64.0 g O₂ 1 mole 2 moles Al₂O₃ 102 g
----------------- x -------------- x ------------------------ x ------------- = 136 g Al₂O₃
32 g 3 moles O₂ 1 mole

Answer 2

Step-by-step explanation:

Based of the fact that you were given 2 masses I would assume this to be a limiting reagent question. However amount on the left side both equal 2. Ignoring limiting reagents and focusing on just O2 the steps would be:

1. Make sure the equation is balanced ( already given)

2- Use given values to find the mols of O2 (mass/molar mass)

3. Mols are conserved but due to the coefficients the molar value from O2 must be divided by three and multiplied by 2 to ensure proper ratios

4. Using that amount the mass can derived using amount/molar mass

5. Use proper significant digits and units(3 in this case)