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.

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

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

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.