• DATA GIVEN

Mass of FeCl3 = 90.0g
Mass of H2S = 52.0g
i. The limiting reactant
ii. Mass of HCl produced
iii. Mass of excess reactant remain after reaction
Equation for the reaction
2FeCl3 + 3H2S F2S3 + 6HCl

• i/ The limiting reactant

To convert mass into mole
Mole for FeCl3 = mass /Molar mass
= 90.0g /162.5
= 0.5538 mole
Mole ratio = mole /Mole coefficient
= 0.5538 mole/2
= 0.2769
Mole of H2S = mass /Molar mass
= 1.529 mole /3 mole
= 0.509

The limiting reactant is FeCl3
• ii/ To find the excess mass of HCL produced

2 moles of FeCl3 Ξ 6 moles of HCl
0.5538 mole of FeCl3 Ξ ?? of HCl
= (0.5538 × 6) moles2 /2 mole
= 1.6614 mole of HCl
= mass of HCl produced
Mass = mole × molar mass
= 1.6614 mole × 36.5g/mole
= 60.6411g
60.6411g of HCl was produced.
• iii/ calculate the mass of H2S remain

Mass reacted from mole equivalent
2moles of FeCl3 Ξ 3 moles of H2S
0.5538 moles of FeCl3 Ξ ??
= 0.5538 mole × 3 moles /2 moles
= 0.8307 mole of H2S
To convert 0.8307 moles into mass
Mass = mole × molar mass
= 0.8307 mole × 34g/mole
= 28.2438g
Mass remain of H2S = mass of origin – mass reacted
= 52.0g – 28.2438g
= 23.7562g of H2S remain unreacted