The Bhopal disaster

Question

Use the following reaction to determine how much of each product would be released if 42 000 kg (42 tonnes) of methyl isocyanate reacted with excess water:

methyl isocyanate + water → 1,3-dimethyl urea + carbon dioxide

2C2 H3 NO + H2 O → C3 H8 N2 O + CO2

Answer 1

molar mass C₂H₃NO = 24 + 3 + 14 + 16 = 57 g / mol

moles C₂H₃NO used in reaction = 42,000 kg x 1000 g / kg x 1 mol / 57 g = 736842 moles C₂H₃NO

Using dimensional analysis and the stoichiometry of the balanced equation, we can find the mass (kg) of each product that will be formed from this amount of C₂H₃NO:

-molar mass C₃H₈N₂O = 88 g / mol

-molar mass CO₂ = 44 g / mol

1. C₃H₈N₂O: 736842 mol C₂H₃NO x 1 mol C₃H₈N₂O / 2 mol C₂H₃NO x 88 g / mol x 1 kg/1000 g = 32421 kg C₃H₈N₂O
2. CO2: 736842 mol C₂H₃NO x 1 mol CO₂ / 2 mol C₂H₃NO x 44 g / mol x 1 kg/1000 g = 16211 kg CO₂