What number will go in the _?_ below to balance the equation?

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

A

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

C

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

$Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

1 year ago

A 6.00 g sample of calcium sulfide is found to contain 3.33 g of calcium. what is the percent by mass of sulfur in the compound?

tankabanditka [31]

2.67 is the hsjshkahsjahsgz hi ajahsghsjahaysjs

8 0

2 years ago

Be sure to answer all parts. There are three different dichloroethylenes (molecular formula C2H2Cl2), which we can designate X,

sertanlavr [38]

Answer:

Explanation:Since the compound X has no net-dipole moment so we can ascertain that this compound is not associated with any polarity.

hence the compound must be overall non-polar. The net dipole moment of compound is zero means that the vector sum of individual dipoles are zero and hence the two individual bond dipoles associated with C-Cl bond must be oriented in the opposite directions with respect to each other.]

So we can propose that compound X must be trans alkene as only in trans compounds the individual bond dipoles cancel each other.

If one isomer of the alkene is trans then the other two isomers may be cis .

Since the two alkenes give the same molecular formula on hydrogenation which means they are quite similar and only slightly different.

The two possibility of cis structures are possible:

in the first way it is possible the one carbon has two chlorine substituents and the carbon has two hydrogens.

Or the other way could be that two chlorine atoms are present on the two carbon atoms in cis manner that is on the same side and two hydrogens are also present on the different carbon atoms in the same manner.

Kindly refer the attachments for the structure of compounds:

7 0

2 years ago

A bottle of antiseptic hydrogen peroxide (H2O2) is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solutio

PolarNik [594]

Answer:

= 12 mL H202

Explanation:

Given that, the concentration of H2O2 is given antiseptic = 3.0 % v/v

It implies that, 3ml H2O2 is present in 100 ml of solution.

Therefore, to calculate the amount of H202 in 400.0 mL bottle of solution;

we have;

(3.0 mL/ 100 mL) × 400 mL

= 12 mL H202

6 0

2 years ago

If 36.0 g of NaOH (MM = 40.00 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the

Kobotan [32]

Answer:

Molarity of NaOH = 1.8 M.

Explanation:

From the question given above, the following data were obtained:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Volume = 500 mL

Molarity of NaOH =?

Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:

Mass of NaOH = 36 g

Molar mass of NaOH = 40 g/mol

Mole of NaOH =?

Mole = mass / molar mass

Mole of NaOH = 36 / 40

Mole of NaOH = 0.9 mole

Next, we shall convert 500 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

500 mL = 500 mL × 1 L / 1000 mL

500 mL = 0.5 L

Finally, we shall determine the molarity of NaOH. This can be obtained as follow:

Mole of NaOH = 0.9 mole

Volume = 0.5 L

Molarity of NaOH =?

Molarity = mole / Volume

Molarity of NaOH = 0.9 / 0.5

Molarity of NaOH = 1.8 M

8 0

1 year ago