Under certain conditions, the substances iron and chlorine combine to form iron(III) chloride. If 31.5 grams of iron and 60.0 grams of chlorine combine to form iron(III) chloride, how many grams of iron(III) chloride must form?

91.5 grams of iron(III)chloride will be formed

Explanation:

Step 1: Data given

Mass of iron = 31.5 grams

atomic mass iron = 55.845 g/mol

Mass of chlorine = 60.0 grams

Molar mass Cl2 = 70.9 g/mol

Step 2: The balanced equation

2Fe + 3Cl2 → 2FeCl3

Step 3: Calculate Moles iron

Moles iron = mass iron / molar mass iron

Moles iron = 31.5 grams / 55.845 g/mol

Moles iron = 0.564 g/mol

Step 4: Calculate moles Cl2

Moles Cl2 = 60.0 grams / 70.9 g/mol

Moles Cl2 = 0.846 moles

Step 5: Calculate moles FeCl3

For 2 moles Fe we need 3 moles Cl2 to produce 2 moles FeCl3

For 0.564 moles iron we'll have 0.564 moles FeCl3

Step 6: Calculate mass FeCl3

Mass FeCl3 = moles FeCl3 * molar mass FeCl3

Mass FeCl3 = 0.564 moles * 162.2 g/mol

Mass FeCl3 = 91.5 grams

91.5 grams of iron(III)chloride will be formed