Calcula el rendimiento de la reacción entre 2 mol de antonimo metálico(masa atomica 121.76 uma) y 5 mol de cloro(Cl2) (masa molar 70.90) para formar 444.6 G de tricloruro de antonimo (SbCl3) (masa molar 228.0 umal la reaccion quimica es la siguiente 4Sb+6Cl2 = 4SbCl3

This is an incomplete question, here is a complete question.

Calculate the percent yield when 2 mol of metal antimony (atomic mass 121.76 amu) and 5 mol of chlorine (molar mass 70.90) to form 444.6 g of antimony trichloride (molar mass 228.0 amu).

The chemical reaction is the following:

Answer : The percent yield of the reaction is, 9.75 %

Explanation :

First we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

From the balanced reaction we conclude that

As, 4 mole of react with 6 mole of

So, 2 moles of react with moles of

From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of

From the reaction, we conclude that

As, 4 mole of react to give 4 mole of

So, 2 mole of react to give 2 mole of

Now we have to calculate the mass of

Molar mass of = 228.0 g/mole

Now we have to calculate the percent yield of the reaction.

Now put all the given values in this formula, we get:

Therefore, the percent yield of the reaction is, 9.75 %