1.Oxygen was discovered by Joseph Priestley in 1774 when he heated mercury (II) oxide, HgO, to decompose it to form its constituent elements. How many moles of mercury (II) oxide are needed to produce 125 g of oxygen? Given:
Unknown:
Mole ratio:
Solution:

2.In a blast furnace, iron (III) oxide is used to produce iron by the following (unbalanced) reaction:
Fe2O3(s) + CO(g)  Fe(s) + CO2(g)
If 4000 g of Fe2O3 is available to react, how many moles of CO are needed?
Given:
Unknown:
Mole ratio:
Solution:

Final

ugh pls help:(

1. 7.81 moles HgO

2. n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Explanation:

How many moles of mercury (II) oxide are needed to produce 125 g of oxygen?

2HgO ==> 2Hg + O2  

125 g O2 x 1 mol O2/32 g x 2 mol HgO / mol O2 = 7.81 moles HgO

If 4000 g of Fe2O3 is available to react, how many moles of CO are needed?

The no. of moles of CO are needed = 75.15 mol.

Fe₂O₃ + 3CO → 2Fe + 3CO₂,

It is clear that 1 mol of Fe₂O₃ reacts with 3 mol of CO to produce 2 mol of Fe and 3 mol of CO₂.

If 4.00 kg Fe₂O₃ are available to react, how many moles of CO are needed?

We need to calculate the no. of moles of 4.00 kg Fe₂O₃:

n = mass/molar mass = (4000 g)/(159.69 g/mol) = 25.05 mol.

Using cross multiplication:

1 mol of Fe₂O₃ need  → 3 mol of CO to react completely, from stichiometry.

25.05 mol of Fe₂O₃ need  → ??? mol of CO to react completely.

The no. of moles of CO are needed = (3 mol)(25.05 mol)/(1 mol) = 75.15 mol.

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6