Question

1. How many moles of mercury(II) oxide, HgO, are needed to produce 125 g of oxygen, O2?
2. What mass of aluminum is produced by the decomposition of 5.0 kg Al2O3?
Please help!

Answer 1

1. We must begin the first question by forming a chemical equation and balancing it.

2HgO -----> 2Hg + O2

The above equation is what the balanced equation would look like if we considered the reaction a decomposition reaction.

Next, we must find the Gram Molecular Weight of O2. This can be done by using the periodic table and multiplying the gram atomic weight of oxygen by 2 since oxygen gas is a molecule.

GMW (Gram Molecular Weight) = 16 * 2 = 32 g/mol

Now lets find out the number of moles that were present in the reaction

n (number of moles) = mass/GMW
n = 125/32 = 3.9

Based on the chemical equation, the number of moles of mercuric oxide will be two times the number of moles of oxygen gas.

3.9 * 2 = 7.8 moles

Therefore, the number of molesof HgO needed to produce 125g of O2 is around 7.8 moles.

2. First convert the kilogams to grams since the periodic table uses mass in grams.

Now lets create the balanced chemical equation

2Al2O3 -------> 4Al + 3O2

Next find out the gram molecular weight of the aluminum oxide by using the periodic table.

GMW of Aluminum Oxide = 27 + 27+ 16 +16 + 16 = 113 g/mol

n = mass/GMW = 5000 / 113 = 44.25 moles

Based on the chemical equation, the number of moles of aluminum is 2 times the number of moles of aluminum oxide.

The number of moles of aluminum produced will be 44.25* 2 = 88.5 moles

Now convert the number of moles to mass.

We know that the Gram Atomic Weight(GAW) of Aluminum from the Periodic Table is around 27g/mol.

Now use the follwing formula

n = mass/GAW

mass = (n)(GAW)

mass = 88.5 * 27 = 2390 g

Therefore the mass of aluminum produced by the decomposition of 5.0 kg Al2O3 is around 2390 g or 2.39 kg.

I hope this helped. Thanks for the great question. Many students find this topic hard to wrap around their minds. But if you keep practicing, at one point you will have no problems.

Answer 2

Sorry for such a long explanation. I just wanted to make sure that you understood the concept.

Answer 3

Thank you!