2.30 g of ethanol were mixed with aqueous acidified potassium dichromate 6. The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%.
What mass of product was collected?
A) 1.54g B) 1.61g C)2.10g D) 3.14g

Question

2.30 g of ethanol were mixed with aqueous acidified potassium dichromate 6. The desired product was collected by immediate distillation under gentle warming. The yield of product was 70.0%.
What mass of product was collected?
A) 1.54g  B) 1.61g  C)2.10g  D) 3.14g

anonymous · Answer

http://en.wikipedia.org/wiki/Potassium_dichromate#Ethanol_determination $\text{CH}_3\text{CH}_2\text{OH}+2[\text{O}]\longrightarrow\text{CH}_3\text{COOH}+\text{H}_2\text{O}$ The molar mass of ethanol is trivial to compute: $M(\text{CH}_3\text{CH}_2\text{OH})=6M(\text{H})+2M(\text{C})+M(\text{O})=46.08\text{ g/mol}$... hence we have $2.30\text{ g}/46.08\text{ g/mol}\approx0.05\text{ mol}$ of ethanol. This yields an equal number of moles of acetate under 100% yield. The molar mass of acetate is just $M(\text{CH}_3\text{COOH})=4M(\text{H})+2M(\text{C})+2M(\text{O})=60.06\text{ g/mol}$... hence we end up with $0.05\text{ mol}\times60.06\text{ g/mol}=3.00\text{ g}$ of acetate with ideal yield. To factor in the fact yield is only 70%, compute $70\%\times3.00\text{ g}=2.10\text{ g}$

anonymous · Answer

thanks