Home » Chemistry » Avogadro’s law

### Reasoning

(from Modern Chemistry, Davis, HRW)

### As shown below, the coefficients in a chemical reaction involving gases indicate the relative numbers of molecules, the relative numbers of moles, and the relative volumes. ## As an equation

### or ## Example problems

### V1 / n1 = V2 / n2

2.00 L / 0.500 mol = 2.70 L / x

x = 0.675 mol

### 0.675 mol – 0.500 mol = 0.175 mol

0.175 mol x 4.00 g/mol = 0.7 grams of He added

### We can perform a calculation using Avogadro’s Law:

V1 / n1 = V2 / n2

Let’s assign V1 to be 1 L and V2 will be our unknown.

Let us assign 1 mole for the amount of neon gas and assign it to be n1.

The mass of argon now added is exactly equal to the neon, but argon has a higher gram-atomic weight (molar mass) than neon. Therefore less than 1 mole of Ar will be added. Let us use 1.5 mol for the total moles in the balloon (which will be n2) after the Ar is added. (I picked 1.5 because neon weighs about 20 g/mol and argon weighs about 40 g/mol.)

1 / 1 = x / 1.5

x = 1.5

### V1 / n1 = V2 / n2

 5.120 L 18.10 L –––––––– = –––––– 8.500 mol x

x = 30.05 mol <— total moles, not the moles added

30.05 – 8.500 = 21.55 mol (to four sig figs)