Example 1
What volume of chlorine at STP reacts with 0.500 mole Fe, and what mass of FeCl3 will be produced?
Given: 0.500 mole Fe,
RTF: Volume of Cl2(g) reacted; mass of FeCl3 produced.
Solution
Step 1: Write the balanced equation:
2Fe(s) + 3Cl2 (g) → 2FeCl3(s)
Step 2: Write the reacting ratios in moles:
2Fe(s) + 3Cl2 (g) → 2FeCl3(s)
Step 3: Determine the volume of chlorine needed:
In general, if aA + bB → cC + dD then amount of B needed = xb/a moles.
Quantity of Cl2 needed = xb/a moles = 0.500 × 3/2 = 0.75 mol Cl2 × 22.4 dm3/mol Cl2
= 16.8 dm3 of chlorine at STP.
Notice that to change 0.75 moles of Cl2 to volume at STP we had to multiply it by the appropriate conversion factor (22.4 dm3/mol).
Step 4: Determine the mass of FeCl3(s) produced:
Quantity of FeCl3(s) produced = xA × (cC/aA) = x.c/a moles (of C).
= 0.500 × (2/2) moles of FeCl3(s) = 0.500 moles of FeCl3(s)
= 0.500 moles of FeCl3(s) × 162.5 g/mol FeCl3(s) = 81.2 g
Again, to find the mass of FeCl3(s) we had to change moles to grams by multiplying by the conversion factor 162.5 g/mol.
