Combustion of gaseous dimethyl ether

Atomic mass of Carbon= 12.01
Atomic mass of Hydrogen= 1.008
Atomic mass of Oxygen= 16
Therefore, Molar mass of dimethyl ether = 46.07g
(as 2 x (12.01+ 3 x 1.008) + 16 = 46.07)

CH3 -O- CH3 (g) + 3O2 (g) --> 2CO2(g) + 3H2O(g)

125 g of dimethyl ether x (1 mol CH3 -O- CH3 / 46.07 g CH3 -O- CH3) = 2.713 mol of dimethyl ether

Molar ratio: 3 mol of oxygen/mol of dimethyl ether

2.713 mol of dimethyl ether x (3 mol oxygen / 1 mol dimethyl ether) = 8.139 mol of oxygen

Therefore to burn 125 g of dimethyl ether gas 2.713 mol of oxygen is required.

From the ideal gas equation
P/RT = rho /M
rho /M = 1.81 g/l / 46.07= 0.039288

Therefore P/RT = 0.039288

For Oxygen
PV = nRT
Or V = n RT /P
n= 8.139 mol
P/RT = 0.039288
Therefore V= 8.139/ 0.039288 = 207 liters

Answer: Volume of Oxygen required= 207 liters

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