.5 mL of bromobutane (MW= 137, density= 1.27 g/mL) was caused to react with exce

Question

.5 mL of bromobutane (MW= 137, density= 1.27 g/mL) was caused to react with excess sodium ethoxide (MW= 68.1) to afford ethoxy butane. What's the maximum yield of product that can be expected in grams?

Explanation / Answer

CH3CH2CH2CH2Br +CH3CH2ONa ->CH3CH2CH2CH2OCH2CH3+ NaBr 0.5mL of bromobutane - converts to molesDensity =mass/volume 1.27 = mass/0.5 Mass of bromobutane = 0.635g n(bromobutane) = 0.635/137 = 0.004635mol n(ethoxy butane) = n(bromobutane) = 0.004635mol n(ethoxy butane) = m/M m(ethoxybutane) = nM = 0.004635 x 68.1 =0.316g Hope this helps!

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