If I have 0.0819 moles of one reactant and 0.6172 moles of the other reactant and...

If I have 0.0819 moles of one reactant and 0.6172 moles of the other reactant and the stoichiometry of the reaction is 1:1:1 how many moles of the product am I supposed to get?

Additional info: the raction 10g of benzoic acid and 25ml of methanol to produce methyl benzoate. I got 9.1ml of product. What is the percent yiled? Please show the calculations.

Expert Solution

Lets say you have 0.0819 moles of A and 0.6172 moles of B and they react to produce C as per the reaction:

A + B ---> C

So, according to the reaction stoichiometry,

1 mole A requires 1 mole B for complete reaction

Since you have more than 0.0819 moles of B, so B is in excess thus A is the limiting reagent.

Thus moles of C produced = moles of limiting reagent present = 0.0819

% yield = (Mass obtained/Theoretical mass)*100

In the additional info, you have not provided the density of the solutions used, so it is not possible to show calculations for that.

Hope this helps ! Revert back if you have any queries.

queen_honey_blossom answered 2 years ago

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