From WolfWikis

Stoichiometry problems

Exercises

Exercise 1

Balance the following equations

C₆H₆ + O₂ => CO₂ + H₂O

First start with the element that occurs the least frequently, that's carbon

C₆H₆ + O₂ => 6CO₂ + H₂O

At least we have as many carbons on the right as on the left.
The next is hydrogen

C₆H₆ + O₂ => 6CO₂ + 3H₂O

Now the hydrogens match. Note that you do need to multiply the 3 in front of the H and the ₂ subscript.
Now count your oxygen atoms and make sure none get lost or created out of thin air There are 6x2+3=15 oxygen atoms on the right. On the left you need to divide by two to get O₂ molecules

C₆H₆ + 7.5O₂ => 6CO₂ + 3H₂O

If you think in moles the 7.5 is fine, but not if you think in molecules, so often people would multiply everthing with two to get:

2C₆H₆ + 15O₂ => 12CO₂ + 6H₂O

The next two go much the same

C₆H₁₂ + O₂ => CO₂ + H₂O

C₆H₁₂ + 9O₂ => 6CO₂ + 6H₂O

CS₂ + O₂ => CO₂ + SO₂

CS₂ + 3O₂ => CO₂ + 2SO₂

This one is a bit trickier. First do S and Al and that gives the coefficients of the left.

Al(OH)₃ + SO₂ => Al₂(SO₃)₃ + H₂O

2Al(OH)₃ + 3SO₂ => Al₂(SO₃)₃ + H₂O

Now do H. There are 2 -(OH)₃ groups i.e. 2*3=6 hydrogens on the left. That means the water needs a 3 in front of it to make 6.

2Al(OH)₃ + 3SO₂ => Al₂(SO₃)₃ + 3H₂O

Now we need to check that O will fall in place. On the left the aluminum hydroxide has 2*3=6 O's, the sulfur dioxide has 3*2= 6 for a total of 12. On the right we have 3*3=9 O's in the sulfite plus 3 O's in the water. Also 12!

Exercise

Compute the molar mass for

Al₂O₃

Aluminum oxide: 2x27+3x16=102 [g/mol]

AuCl₃

Gold(III)chloride: 197+3*35.5= 303.5 [g/mol]

Rb₂SeO₄

Rubidium selenate: 2*85.5+ 79+4*16=314 [g/mol]

Ho(OH)₃

Holmium hydroxide 165+3*(16+1)= 216 [g/mol]

C₁₂(OH)₃N₂O₃F₃

I have no idea, I made this up: 12*12+3*17+2*14+3*16+3*19=328 [g/mol]

Exercise 3

I have one gram of the following compounds; how many moles do I have?

Molar masses are in [g/mol]. The question does the opposite: it asks for [mol/g]. This means you need to compute the molar mass and then take its reciprocal

Ca(OH)₂

1 g of Calcium hydroxide is 1/[40+2x17]=1/[74] [mol/g].[g]= 0.0135 [mol]

AuCl

1 g of Gold(I) chloride is 1/[197+35.5]=1/[232.5] [mol/g].[g]= 4.3 [mmol]. Note that if you get too many zeros it is easier to go to milimoles etc.

Rb₂SO₄

1 gram of Rubidium sulfate: 1/[2*85.5+ 32+4*16]= 1/[] [mol/g].[g] = 3.7 [mmol]

HoBr₃

1 gram of Holmium bromide 1/[165+3*(80)]= = 2.4 [mmol]

C₁₂(OH)₃N₂O₃F₃

1 gram of whatever: 1/[12*12+3*17+2*14+3*16+3*19=328] = 3.0 [mmol]

Exercise 4

I add one mole of oxygen in the following reactions and it is exactly enough to burn up the other reactant. How many grams reactant do I need?

Warning: the equations are not balanced!

WARNING: such warnings are not always given! You need to make sure equations are balanced at all times!!!

C₆H₁₄ + O₂ => CO₂ + H₂O

C₆H₁₄ + 9.5O₂ => 6CO₂ + 7H₂O

One mole of oxygen can only react with 1/(9.5)= 0.105 moles of C₆H₁₄.
One way to prove that is to remember that we can always multiply all the coefficients with the same number. I.e. we could divide all of them by 9.5 to make sure the oxygen coefficient is 1:

[1/9.5] C₆H₁₄ + [9.5/9.5]O₂ => [6/9.5]CO₂ + [7/9.5]H₂O

C₅H₁₂ + O₂ => CO₂ + H₂O

C₅H₁₂ + 8O₂ => 5CO₂ + 6H₂O

[1/8] mol of C₅H₁₂ will react with one mole of oxygen

CSH₄ + O₂ => CO₂ + SO₂ + H₂O

CSH₄ + 3O₂ => CO₂ + SO₂ + 2H₂O

Burning 1/3 of a mole of very smelly methyl mercaptane is burnt by one mole of oxygen and produces suffocating sulfur dioxide. Yikes.