2hgo(s) --> 2hg(l) + o2 (g) in a certain reaction 8.74 g of hgo is decomposed, producing 6.42 g of hg. what is the percent yield for the reaction?

Percent yield describes the amount of product produced in a reaction relative of the amount of reactant you started with. In this question, you begin with 8.74 grams of HgO, and dividing this by it’s formula mass of 216.58g (200.59 for Mercury and 15.99 for Oxygen), we find that this amount of HgO is 0.0404 moles.

You end with 6.42 grams of Hg, or 0.032 moles, when divided by its formula weight of 200.59g.

The percent yield is calculated by dividing the number of moles of product by the number of moles of reactant and multiplying by 100, so the percent yield is:

0.032/0.0404 = 0.792 x 100 = 79.2%.

The reaction proceeds at 79.2% efficiency.

Where's the pictures ?

U whay do you need with?