Editing Stoichiometry (section)

== Determining amount of product ==
Stoichiometry can also be used to find the quantity of a product yielded by a reaction. If a piece of solid [[copper]] (Cu) were added to an aqueous solution of [[silver nitrate]] ({{chem2|AgNO3}}), the [[silver]] (Ag) would be replaced in a [[single displacement reaction]] forming aqueous [[copper(II) nitrate]] ({{chem2|Cu(NO3)2}}) and solid silver.  How much silver is produced if 16.00 grams of Cu is added to the solution of excess silver nitrate?

The following steps would be used:
# Write and balance the equation
# Mass to moles: Convert grams of Cu to moles of Cu
# Mole ratio: Convert moles of Cu to moles of Ag produced
# Mole to mass: Convert moles of Ag to grams of Ag produced

The complete balanced equation would be:
: {{chem2|Cu + 2 AgNO3 -> Cu(NO3)2 + 2 Ag}}

For the mass to mole step, the mass of copper (16.00&nbsp;g) would be converted to moles of copper by dividing the mass of copper by its [[molar mass]]: 63.55&nbsp;g/mol.
: <math>\left(\frac{16.00 \mbox{ g Cu}}{1}\right)\left(\frac{1 \mbox{ mol Cu}}{63.55 \mbox{ g Cu}}\right) = 0.2518\ \text{mol Cu}</math>

Now that the amount of Cu in moles (0.2518) is found, we can set up the mole ratio. This is found by looking at the coefficients in the balanced equation: Cu and Ag are in a 1:2 ratio.
: <math>\left(\frac{0.2518 \mbox{ mol Cu}}{1}\right)\left(\frac{2 \mbox{ mol Ag}}{1 \mbox{ mol Cu}}\right) = 0.5036\ \text{mol Ag}</math>

Now that the moles of Ag produced is known to be 0.5036&nbsp;mol, we convert this amount to grams of Ag produced to come to the final answer:
: <math>\left(\frac{0.5036 \mbox{ mol Ag}}{1}\right)\left(\frac{107.87  \mbox{ g Ag}}{1 \mbox{ mol Ag}}\right) = 54.32 \ \text{g Ag}</math>

This set of calculations can be further condensed into a single step:
: <math>m_\mathrm{Ag} = \left(\frac{16.00 \mbox{ g }\mathrm{Cu}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{Cu}}{63.55 \mbox{ g }\mathrm{Cu}}\right)\left(\frac{2 \mbox{ mol }\mathrm{Ag}}{1 \mbox{ mol }\mathrm{Cu}}\right)\left(\frac{107.87 \mbox{ g }\mathrm{Ag}}{1 \mbox{ mol Ag}}\right) = 54.32 \mbox{ g}</math>

=== Further examples ===
For [[propane]] ({{chem2|C3H8}}) reacting with [[oxygen|oxygen gas]] ({{chem2|O2}}), the balanced chemical equation is:
: {{chem2|C3H8 + 5 O2 -> 3 CO2 + 4 H2O}}

The mass of water formed if 120&nbsp;g of propane ({{chem2|C3H8}}) is burned in excess oxygen is then
: <math>m_\mathrm{H_2O} = \left(\frac{120. \mbox{ g }\mathrm{C_3H_8}}{1}\right)\left(\frac{1 \mbox{ mol }\mathrm{C_3H_8}}{44.09 \mbox{ g }\mathrm{C_3H_8}}\right)\left(\frac{4 \mbox{ mol }\mathrm{H_2O}}{1 \mbox{ mol }\mathrm{C_3H_8}}\right)\left(\frac{18.02 \mbox{ g }\mathrm{H_2O}}{1 \mbox{ mol }\mathrm{H_2O}}\right) = 196 \mbox{ g}</math>