Question

2 Fe(s) + 3 Cl2 (g) → 2 FeCl3 (s) a) Beginning with 10.0 g of iron, what mass of Cl2 (g) is required for complete reaction?

Answer 1

The atomic mass of iron is 55.845 g/mol, so 10.0 g is equal to
`10.0/55.845=0.1790670606142 \text{ mol}`.

From the equation, we know that for every 2 moles of iron consumed, 3 moles of chlorine are consumed.

This means we need
`3\left((0.1790670606142)/(2) \right)=0.2686005909213 \text{ mol}` chlorine.

Chlorine has an atomic mass of 35.45 g/mol, so
`\text{Cl}_(2)` has a formula mass of 35.45(2)=70.9 g/mol.

Thus, the mass of chlorine needed is (0.2686005909213)(70.9), which is about 19.0 g (to 3 sf)