What volume of water is produced when 38.5 g of ethanol reacts with oxygen at 500°C at 1.75 atm? CH3CH2OH(g) + 3 O2(g)→ 2 CO2(g) + 3 H2O(g)

The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

The given chemical equation follows:

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = of water

The ideal gas equation is given as:

.......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank =

Putting values in equation 2, we get:

Hence, the volume of oxygen gas is 91.4 L.

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