Stoichiometry— In a chemical reaction, the total weight of reactants equals the total weight of products

due to the Law of Conservation of Mass. Stoichiometry uses this law to relate the weights of reactants
and products using a balanced chemical equation. This equation provides the ratio of reactants to
products in terms of moles, which can be converted to weight using molar mass.

In the equation:

CH4+2O2→CO2+2H2O

 CH4 (methane) and O2 (oxygen) are the things we start with. These are called reactants.
 CO2 (carbon dioxide) and H2O (water) are what we end up with. These are called products.
 The numbers in front of the molecules tell us how many of each we need:
 We need 1 molecule of methane and 2 molecules of oxygen to start.
 The reaction will produce 1 molecule of carbon dioxide and 2 molecules of water.

This is a complete combustion reaction, which means a fuel (here, methane) burns in oxygen to produce
carbon dioxide and water. The total weight of the reactants equals the total weight of the products,
which is a principle known as the Law of Conservation of Mass. This law states that matter cannot be
created or destroyed, only transformed.

CONVERTING GRAMS TO MOLES

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between
reactants and products in chemical reactions. It’s not just about balancing chemical equations but
also involves conversions between different units of measurement.

For instance, you might need to convert the mass of a substance (in grams) to the number of
moles, using the molar mass as a conversion factor. The molar mass is calculated based on the
atomic masses of its constituent elements.

In your example, if you have 2.00 g of NaCl (sodium chloride) and want to find out how many
moles that is, you would use the molar mass of NaCl as a conversion factor. The molar mass of
NaCl is approximately 58.44 g/mol, which is calculated based on the atomic masses of sodium
(Na) and chlorine (Cl).

So, for 2.00 g of NaCl, the calculation would be:

Given:

Mass of NaCl = 2.00 g

Molar mass of NaCl = 58.44 g/mol

2.00 g
n=
58.44 g /mol

This means there are 0.0342 moles of NaCl in a 2.00 g sample.

MOLAR PROPORTION

Stoichiometry is indeed used to balance chemical equations, which is often referred to as

For example, the reaction of hydrogen (H2) and oxygen (O2) to form water (H2O) can be
represented by the following balanced chemical equation:

2H2+O2→2H2O

This equation tells us that two molecules of hydrogen react with one molecule of oxygen to
produce two molecules of water. This is the 2:1:2 ratio of hydrogen, oxygen, and water
molecules you mentioned.

In the case of the combustion of methanol (CH3OH), the balanced chemical equation is:

2CH3OH+3O2→2CO2+4H2O

This equation tells us that two molecules of methanol react with three molecules of oxygen to
produce two molecules of carbon dioxide and four molecules of water.

If you have 0.27 moles of methanol and want to find out how many moles of water will be
produced, you would use the molar ratio of methanol to water from the balanced chemical
equation, which is 2:4 or 1:2.

So, for 0.27 moles of methanol, the calculation would be:

a. Identify the given quantity: The given is 0.27 mol of CH3OH (methanol).
b. Identify the molar ratio: The molar ratio is taken from the balanced chemical equation. In
this case, it’s 4 mol of H2O (water) for every 2 mol of CH3OH. This ratio is represented
as 4 mol H2O / 2 mol CH3OH.
 c. Perform the conversion: Multiply the given moles of CH3OH by the molar ratio to find
the moles of H2O. The CH3OH units cancel out, leaving you with moles of H2O.

Moles of H 20 4 mol H 20
Moles of H 20=Moles of CH 3 OH × =0.27 mol × =0.54 mol H 2 O
Moles of CH 3 OH 2 mol CH 3 OH

This means that the combustion of 0.27 moles of methanol would produce 0.54 moles of water.

DETERMINING AMOUNT OF PRODUCT

The reaction you’re referring to is a single displacement reaction where copper (Cu) displaces
silver (Ag) in silver nitrate (AgNO3) to form copper(II) nitrate (Cu(NO3)2) and silver (Ag). The
balanced chemical equation for this reaction is:

Cu+2AgNO3→Cu(NO3)2+2Ag

This equation tells us that one mole of copper reacts with two moles of silver nitrate to produce
one mole of copper(II) nitrate and two moles of silver.

If you have 16.00 grams of copper and want to find out how many grams of silver will be
produced, you would first need to convert the mass of copper to moles using its molar mass,
which is approximately 63.55 g/mol.

So, for 16.00 g of copper, the calculation would be:

Mass of Cu 16.00 g
Moles of Cu= = =0.2518 mol
Molar Mass of Cu 63.55 g /mol

Then, you would use the molar ratio of copper to silver from the balanced chemical equation,
which is 1:2, to find out how many moles of silver will be produced:

Moles of Ag 2 mol Ag
Moles of Ag=Moles of Cu × =0.2518 mol × =0.5036 mol Ag
Moles of Cu 1 mol Cu

Finally, you would convert the moles of silver to grams using its molar mass, which is
approximately 107.87 g/mol:

g
Mass of Ag=Moles of Ag× Molar Mass of Ag=0.504 mol ×107.87 =54.32
mol
So, if you add 16.00 grams of copper to an excess solution of silver nitrate, you would produce
approximately 54.32 grams of silver.

STOICHIOMETRIC RATIO

Stoichiometry is a branch of chemistry that deals with the quantitative relationships between the
reactants and products in a chemical reaction. It allows us to predict how much of a reactant is
needed or product is formed in a chemical reaction.

The example you provided is the thermite reaction, which is a type of redox reaction. Here’s the
balanced chemical equation:

Fe2O3+2Al→Al2O3+2Fe

This equation tells us that one mole of iron(III) oxide (Fe2O3) reacts with two moles of
aluminum (Al) to produce one mole of aluminum oxide (Al2O3) and two moles of iron (Fe).

The image you provided shows a stoichiometric calculation. It calculates the mass
of aluminum needed to completely react with 85.0 g of Fe2O3 in the thermite
reaction. The calculation is as follows:

This means that 28.7 g of aluminum is needed to completely react with 85.0 g of
iron(III) oxide in the thermite reaction. This ensures that there are no leftover
reactants when the reaction takes place. This is the principle of stoichiometry.

GAS STOICHIOMETRY
The calculation involves converting the given mass of a reactant to moles using its molar mass, then using
the stoichiometric coefficients from the balanced chemical equation to find the number of moles of the
gas produced, and finally converting the moles of gas to volume using the ideal gas law.

The image you provided explains the concept of gas stoichiometry, which is used to calculate the
unknown volume or mass of a gaseous product or reactant in a chemical reaction.

For example, if we wanted to calculate the volume of gaseous (nitrogen dioxide)NO2 produced
from the combustion of 100 g of (ammonia)NH3, we would use the reaction:

4 NH3(g)+7 O2(g)→4 NO2(g)+6 H2O(l)

This reaction tells us that 4 moles of (ammonia)NH3 react with 7 moles of (oxygen)O2 to
produce 4 moles of (nitrogen dioxide)NO2 and 6 moles of (water)H2O.
Using this stoichiometric relationship, we can calculate the volume of NO2 produced from a
given amount of NH3. This involves converting the mass of NH3 to moles, using the molar mass
of NH3, and then using the stoichiometric ratio from the balanced equation to find the number of
moles of NO2 produced. The volume of NO2 can then be calculated using the ideal gas law,
given the temperature and pressure conditions.

The image you provided shows a calculation that converts the mass of a substance (in this case,
ammonia, NH₃) to moles. This is a common type of calculation in chemistry known as mole
conversion.

Here’s the breakdown of the calculation:

100g NH₃×17.034g NH₃1mol NH₃=5.871mol NH₃

This calculation is saying that if you have 100 grams of ammonia (NH₃), it is equivalent to 5.871
moles of ammonia. The number 17.034 in the denominator is the molar mass of ammonia, which
is the mass of one mole of ammonia.

This type of calculation is important in stoichiometry, which is the study of the quantitative
relationships between reactants and products in a chemical reaction. It allows chemists to predict
how much of a reactant is needed or how much product will be formed in a chemical reaction.