Chapter 1 The Early Experiments
Example 1.
How much AlO 2can be produced from 10.0 mol Al and 9.0 mol O 3? 2The balanced chemical equation is 4Al + 3O→ 22AlO 2. 3
To find the limiting reactant, we must determine how much AlO 2can be made from each 3reactant.10.0 mol Al×232 mol Al O4 mol Al232= 5.0 mol Al O9.0 mol O×2322 mol Al O3 mol O23= 6.0 mol Al OThere is enough Al to make 5.0 mol AlO 2, and enough O 3to make 6.0 mol Al 2O 2. 3
Therefore, Al is the limiting reactant and 5.0 mol AlO 2is produced. 3How many moles of excess reactant remain? The number of moles of Oconsumed in the reaction 2with 10.0 mol Al is×223 mol O10.0 mol Al= 7.5 mol O4 mol Al7.5 of the 9.0 moles of Oreact, so there are 9.0 – 7.5 = 1.5 mol O 2left over. 2Dalton’s atomic theory proved successful in pr
edicting the experimental results of his
time and became the accepted way to think about matter. His ideas on atomic masses also proved to be very useful, although some of
the atomic masses had to be changed as new
data became available. All in all, chemists
of the day were quite comfortable with the idea
that the smallest unit of matter was the ‘billiard
ball’ atom proposed by Dalton. Then, near
the end of the 19th century, new and more sophi
sticated experiments were performed that
caused scientists to change their view. Howeve
r, before we examine these experiments, we
need to understand the roles of energy and charge in the study of chemistry.
1.ENERGY
Energy plays an important role throughout chem
istry. Indeed, there is hardly a chapter in
this text in which energy considerations are not required. In simple terms, energy is the capacity to move something. The energy of a s
ubstance is the sum of its kinetic energy and
its potential energy.
Kinetic energy (KE)
is energy of motion; an object that is moving
has the capacity to make another object move simply by colliding with it. The kinetic energy of a particle of mass m moving with a velocity v is given in Equation 1.1.
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