To prepare you for your next test we'll go over the common conversion factors
used in enthalpy stoichiometry, a common test question and at the end
you'll get to try a question out on your own. Let's get started with our common
conversion factors. Our first conversion factor used is molar mass. Remember molar
mass allows us to convert from grams to moles or from moles to grams. The new
conversion factor we will be using is the enthalpy change this is extremely
important! Now let's say we have this balanced equation with our enthalpy
change, there are three different conversion factors we have for this
example because our enthalpy change has the units of kilojoules per mole. If we
were looking for Sulfur then we would place the number of moles for sulfur
which is two moles as shown on our balanced equation and we'll leave the
kilojoules on top. What keeps changing is the number of moles on the bottom based
on whatever element you are looking for. This page is part of my 50 page thermochemistry
guide which has everything you need to know for thermochemistry and
enthalpy stoichiometry, the link is in the description box along with other
helpful resources. Now for this common test question, we will find the balanced
equation heat and the mass let's start with finding the balanced chemical
equation for Part A. Our question reads a thousand two hundred and four kilojoules
of heat energy is released when magnesium metal combines with oxygen.
This part translates into magnesium being our first reactant combines meaning
plus and our second reactant is oxygen. Now we do not have just one oxygen
instead oxygen has a two subscript this is because oxygen is a type of diatomic
molecule. Diatomic molecules are atoms that exist in pairs
when we combine magnesium and oxygen together we are only looking at one
oxygen so ignore the subscript of 2 and we will identify our charges. Magnesium
has a +2 charge and oxygen has a -2 charge, since the charges are
the same number and opposite signs we do not need any
scripts so our product is magnesium oxide. Balance the chemical equation by
separating each element and counting how much of each element there is on either
side. On the reactant side we have one magnesium and two oxygens and on the
product side we have one magnesium and one oxygen. We need to balance out the
oxygens so we'll add a two in front of our product and we now have two
magnesiums and two oxygens on the product side. So now the magnesium needs
a 2 in front on the reactant side and this is our balanced equation. This
question also states heat energy is released meaning our enthalpy change is
released which means it's negative. Note: if heat or energy is released then our
enthalpy change is negative, if heat or energy is absorbed that our enthalpy
change is positive. We'll use everything we found in Part A
for Part B, which states calculate the amount of heat transferred when
2.40 grams of magnesium reacts. So we are given the mass of magnesium and
we're finding heat, we'll use the main conversion factors we
discussed at the beginning of this video which were molar mass and enthalpy
change. To get from grams to kilojoules which is what heat is measured in, we
will go from the grams of magnesium to the moles of magnesium to kilojoules.
Always start with what we are given which was the 2.40 grams of
magnesium to convert grams to moles we will use the molar mass of magnesium
which is found on the periodic table. Align grams across from each other so
they can cancel and now we are at moles of magnesium. Next we need to convert to
kilojoules using the given enthalpy change found in Part A our units for
enthalpy change are really kilojoules per mole. So since we are looking at
magnesium our conversion factor will include the number of moles we have for
magnesium in our balanced equation which is two. We'll place two moles of magnesium
on the bottom since we want to align our units of moles to cancel out and on top
we will have our kilojoules since this is what we're looking for. Multiply each
number straight across then divide by the numbers multiplied and round to
three sig figs since our given value had three sig
figs. Part C states how many grams of magnesium oxide are produced during an
enthalpy change of -96.0 kilojoules. So we are given our enthalpy
change of -96.0 kilojoules and asked to find the grams of magnesium
oxide. We'll start with the given unit of kilojoules and make our way to moles of
magnesium oxide then using the molar mass of magnesium oxide we'll find the
grams our setup will start with our given value of -96.0
kilojoules and we will use the enthalpy change as our conversion factor. This
time it's flipped because we always align our units so they can cancel out
so kilojoules goes on the bottom and two moles of magnesium oxide goes on top. We
found this from our balanced equation, next we will convert our moles of
magnesium oxide to grams so we'll use the molar mass of magnesium oxide. To
find the molar mass of magnesium oxide we will add up each elements individual
mass and get 40.31 grams per mole. The moles cancel and we get 6.43
grams of magnesium oxide. Try this question out on your own and we'll
go over the step-by-step answers in this next video and if you'd like more
chemistry help make sure to check the description box for my recommended
resources.