Red mercury (ii) oxide decomposes to form mercury metal and oxygen gas according to the following equation: 2hgo (s) 2hg (l) + o2 (g). if 4.00 moles of hgo decompose to form 1.50 moles of o2 and 603 g of hg, what is the percent yield of this reaction?

hey there!:

2HgO (s) =>  2Hg (l) + O2 (g)

2 moles of HgO decompose to form 2 moles of Hg and 1 mole of O2 according to the reaction mentioned in the question.

So 4.00 moles of HgO must give 4 moles of Hg and 2 moles of O2 theoretically.

603 g of Hg = 603 / 200.6 = 3 moles

Percent yield = ( actual yield / theoretical yield) * 100

= ( 3/4) * 100

= 75 %

Hope this helps!

i believe you are right for all except the 3rd ( look at sqdancefan his answer is much better)

frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa