How much oxygen is needed to burn 3.9 g of benzene?

1. Let’s write down the reaction equation:

2C6H6 + 15O2 = 12CO2 + 6H2O.

2. Find the amount of benzene:

n (C6H6) = m (C6H6) / M (C6H6) = 3.9 g / 78 g / mol = 0.05 mol.

3. Using the equation, we find the amount of oxygen and its volume (Vm is the molar volume equal to 22.4 mol / l)

6n (C6H6) = 15n (O2)

n (O2) = 6n (C6H6) / 15 = 6 * 0.05 mol / 15 = 0.02 mol.

V (O2) = n (O2) * Vm = 0.02 mol * 22.4 mol / L = 0.448 L.

Answer: V (O2) = 0.448 l.




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