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Combustion of gaseous dimethyl ether

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What volume of O2 (g) is consumed in the combustion of 125g of gaseous dimethyl ether (CH3)2) if both gas volumes are measured at the temperature and pressure at which the density of dimethyl ether is 1.81g/l?

https://brainmass.com/chemistry/gas-laws/combustion-of-gaseous-dimethyl-ether-32971

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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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