Question

Question #b8b2b

Answer 1

This new yield will allow for the production of `"8.03 g"` of ethyl butyrate.

Since this question builds upon what was already solved here

http://socratic.org/questions/given-7-80g-of-butanoic-acid-and-excess-ethanol-how-many-grams-of-ethyl-butyrate#131327

I'll skip the balanced chemical equation and the theoretical yield and just show you how to get the new mass produced.

So, the theoretical yield for 7.80 g of butanoic acid was calculated at 10.3 g. When the reaction had a percent yield of 54.9%, only 5,65 g were produced.

It is safe to assume that the new yield of 78.0% will allow for the production of more ethyl butyrate. To determine this new mass, use the definition of the percent yield

`"% yield" = "actual yield"/"theoretical yield" * 100`

Reearranging this equation to get the actual yield will result in

`"actual yield" = ("% yield" * "theoretical yield")/"100"`

`"actual yield" = (78.0 * "10.3 g")/"100" = "8.034 g"`

Rounded to three sig figs, the mass of ethyl butyrate produced will be

`m_("ethyl butyrate") = "8.03 g"`

Answer 2

We first calculate the molecular masses of butanoic acid and ethyl butanoate.

Butanoic acid `88.11` g/Mol
Ethyl butanoate `116.16` g/Mol

From the butanoic acid `78%=0.78*7.80=6.084g` is used

Now we work with ratios, because every 88.11g used will yield 116.16g

`6.084/88.11=x/116.16->x=(116.16*6.084)/88.11~~8.02g`

Answer: 8.02 grams of ethyl butanoate will be formed.