Question

Al(s) + O2(g) Al2O3(s) (a) Consider the unbalanced equation above. How many grams of O2 are required to react with 49.0 g of aluminum? Use at least as many significant figures in your molar masses as in the data given.

Answer 1

We need 43.58 grams of O2 to react with 49.0 grams of aluminium

Step-by-step explanation:

Step 1: Data given

Mass of aluminium = 49.0 grams

Molar mass of aluminium = 26.98 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)

Step 3: Calculate moles aluminium

Moles aluminium = mass aluminium / molar mass aluminium

Moles aluminium = 49.0 grams / 26.98 g/mol

Moles aluminium = 1.816 moles

Step 4: Calculate moles O2 needed

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

For 1.816 moles aluminium we need 3/4 * 1.816 = 1.362 moles O2

Step 5: Calculate mass O2

Mass O2 = moles O2 * molar mass O2

Mass O2 = 1.362 moles * 32.0 g/mol

Mass O2 = 43.58 grams

We need 43.58 grams of O2 to react with 49.0 grams of aluminium