Question

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

Answer 1

Answer:

136 g Al₂O₃

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

Answer:

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)