1) In concentrated sulfuric acid, ethanol (CH 3 CH 2 OH) forms diethyl ether (CH

Question

1) In concentrated sulfuric acid, ethanol (CH3CH2OH) forms diethyl ether (CH3CH2OCH2CH3) and water according to the reaction

2 CH3CH2OH ? CH3CH2OCH2CH3 + H2O

In a side reaction, some ethanol forms ethylene (CH2CH2) and water

CH3CH2OH ? CH2CH2 + H2O

In a given experiment, 50 g of ethanol yielded 33.9 g of diethyl ether and 50% of the ethanol that did not produce diethyl ether reacted according to the side reaction to form ethylene. What mass of ethylene was produced?

A) 1.2 g           B) 2.4 g           C) 5.6 g           D) 10.2 g         E) 15.9 g

Explanation / Answer

molar mass of diethyl ether, ethanol and ethylene are respectively 74.1 g/mol, 46.1 g/mol and 28.1 g/mol.

mole of diethyl ether formed = 33.9 / 74.1 = 0.457 mol

mole of ethanol consumed for diethyl ether = 2* mol of ethanol

                                                                               = 0.914 mol

mass of ethanol consumed for diethyl ether formation = 0.914 * 46.1 = 42.1 g

ethnaol that didnot produced diethyl ether = 50 - 42.1 = 7.9 g

50 of ethnaol that didnot produced diethyl ether = 7.9 * 50 /100 = 3.95 g

mol of ethanol that produced CH2CH2 = 3.95 / 46.1 = 0.0857 mol

mole of ethanol = mole of CH2CH2 = 0.0857 mol

mass of CH2CH2 produced = 28.1 * 0.0857 = 2.4 g

Thus correct answer is B) 2.4 g

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