Question

The expression of the theoretical yield (TY) in function of limiting reagent (LR) of a reaction is as follows: TY = ideal mole ratio of (target product / LR) x #mol(LR) x MW(target product) The ideal mole ratio is the one provided by the equation of the reaction. If a reaction uses (4.50x10^0) g of aniline and 1.25 times as many mL of acetic anhydride as the number of grams of aniline, what is the theoreticl yiled of acetanilide (MW = 135.17 g/mol) in the reaction?

Answer 1

Answer: The theoretical yield of acetanilide is 6.5 grams.

Step-by-step explanation:

To calculate the number of moles, we use the equation:

`\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}` .....(1)

• For aniline:

Given mass of aniline =
`4.50* 10^0=4.50g` (We know that:
`10^0=1` )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

`\text{Moles of aniline}=(4.50g)/(93.13g/mol)=0.048mol`

• For acetic anhydride:

To calculate the mass of acetic anhydride, we use the equation:

`\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}`

Volume of acetic anhydride =
`(1.25* \text{Mass of aniline})=1.25* 4.50=5.625mL`

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

`1.08g/mL=\frac{\text{Mass of acetic anhydride}}{5.625mL}\\\\\text{Mass of acetic anhydride}=(1.08g/mL* 5.625mL)=6.08g`

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

`\text{Moles of acetic anhydride}=(6.08g)/(102.1g/mol)=0.06mol`

The chemical equation for the reaction of aniline and acetic anhydride follows:

`C_6H_5NH_2+CH_3COOCOCH_3\rightarrow C_6H_5NHCOCH_3+CH_3COOH`

By Stoichiometry of the reaction:

1 mole of aniline reacts with 1 mole of acetic anhydride

So, 0.048 moles of aniline will react with =
`(1)/(1)* 0.048=0.048mol` of acetic anhydride

As, given amount of acetic anhydride is more than the required amount. So, it is considered as an excess reagent.

Thus, aniline is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of aniline produces 1 mole of acetanilide

So, 0.048 moles of aniline will produce =
`(1)/(1)* 0.048=0.048mol` of acetanilide

Now, calculating the theoretical yield of acetanilide by using equation 1:

Moles of acetanilide = 0.048 moles

Molar mass of acetanilide = 135.17 g/mol

Putting values in equation 1, we get:

`0.048mol=\frac{\text{Mass of acetanilide}}{135.17g/mol}\\\\\text{Mass of acetanilide}=(0.048mol* 135.17g/mol)=6.5g`

Hence, the theoretical yield of acetanilide is 6.5 grams.