How many grams of ethyl alcohol would be needed to produce 104.5 g of diethyl ether if the percent yield of reaction is 83.4 % ?

The reaction to convert ethyl alcohol to diethyl ether with the use of concentrated sulfuric acid as a catalyst at 140°C, and with excess ethanol. The reaction is written below:

2 CH₃CH₂OH  →  CH₃CH₂OCH₂CH₃  + H₂O

The molar mass of diethyl ether is 74 g/mol while that of ethanol is 46 g/mol. The solution is as follows:

(104.5 g diethyl ether)*(0.834)*(1 mol diethyl ether/74 g)(2 mol ethanol/1 mol diethyl ether)*(46 g ethanol/mol) = 108.35 g ethanol needed