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Given the following reaction: 2NH3(g) + 3Cl2(g) → N2(g) + 6HCl(g)? How many grams of chlorine must react to produce 16 L of nitrogen gas at 1.2 ATM and 23 °C?​

Respuesta :

solution:

Consider the N₂ gas produced:

Pressure, P = 1.2 atm

Volume, V = 16 L

Gas constant, R = 0.08206 L atm / (mol K)

Temperature, T = (273 + 23) K = 296 K

Gas law: PV = nRT

Then, n = PV/(RT)

No. of moles of N₂, n

= 1.2 × 16 / (0.08206 × 300)

= 0.79 mol

According to the given equation, mole ratio Cl₂ :

N₂ = 3 : 1

No. of moles of Cl₂ needed 

= (0.79 mol) × 3

= 2.37 mol

Mass of Cl₂ needed 

= (2.37 mol) × (35.5 × 2 g/mol) 

= 168 g