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?

The theoretical yield of acetanilide is 6.5 grams.

Explanation:

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

     .....(1)

Given mass of aniline =      (We know that:   )

Molar mass of aniline = 93.13 g/mol

Putting values in equation 1, we get:

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

Volume of acetic anhydride =

Density of acetic anhydride = 1.08 g/mL

Putting values in above equation:

Given mass of acetic anhydride = 6.08 g

Molar mass of acetic anhydride = 102.1 g/mol

Putting values in equation 1, we get:

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

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 = 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 = 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:

Hence, the theoretical yield of acetanilide is 6.5 grams.

"you start with 15.67 g mgco3-xh2o, when heated and all the water is removed, you are left with 7.58 g mgco3 only.

therefore, you had 15.67 - 7.58 = 8.09g h2o  

moles of each :

7.58g mgco3 = ? moles = 7.58g / 83.4g/mole = 0.09moles  

8 09g h2o = ? moles = 8.09g / 18g/mole = 0.44moles  

molar ratio between the two by dividing both numbers by the smallest number :

0.09 / 0.09 = 1  

0.44 / 0.09 = 4.8  

mgco3-5h2o"

c8h10 is your answerr