Given the following reaction:
\qquad
R1_RATIO === 1 ? "" : R1_RATIOR1
+ R2_RATIO === 1 ? "" : R2_RATIOR2
\rightarrow
P1_RATIO === 1 ? "" : P1_RATIOP1
+ P2_RATIO === 1 ? "" : P2_RATIOP2
How many moles of P1 will be produced from
GIVEN_MASS \text{g} of GIVEN, assuming NOT_GIVEN is available in excess?
\dfrac{GIVEN_MASS \cancel{\text{g}}}{GIVEN_MOLAR_MASS \cancel{\text{g}} / \text{mol}} =
\text{ plural(GIVEN_MOL, "mole")} \text{ OF }GIVEN
[Explain]
First we want to convert the given amount of GIVEN from grams to moles. To do this, we divide
the given amount of GIVEN by the molar mass of GIVEN.
\dfrac{\text{GRAMS_OF }GIVEN}{\text{MOLAR_MASS_OF }GIVEN} = \text{MOLES_OF }GIVEN
To find the molar mass of GIVEN, we look up the atomic weight of each atom in a molecule of
GIVEN in the periodic table and add them together.
In this case, it's GIVEN_MOLAR_MASS \text{g/mol}.
Dividing the given GIVEN_MASS \text{g} of GIVEN by the molar mass of
GIVEN_MOLAR_MASS \text{g/mol} tells us we're starting with
\text{GIVEN_MOL plural_form(MOLE, GIVEN_MOL)} of GIVEN.
The mole ratio of \dfrac{GIVEN}{P1} in the reaction is
\dfrac{GIVEN_RATIO}{P1_RATIO}.
[Explain]
The reaction is \blue{GIVEN_RATIO}GIVEN
+ R2_RATIOR2 \rightarrow
\red{P1_RATIO}P1
+ P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(GIVEN_RATIO) GIVEN for every
cardinalThrough20(P1_RATIO) P1 molecule.
The reaction is \blue{GIVEN_RATIO}GIVEN
+ R2_RATIOR2 \rightarrow
\red{P1_RATIO}P1
+ P2_RATIOP2.
The coefficients in front of each molecule tell us in what ratios the molecules react. In this case
cardinalThrough20(GIVEN_RATIO) GIVEN for every
cardinalThrough20(P1_RATIO) P1 molecules.
\qquad
\dfrac{GIVEN}{P1} = \dfrac{GIVEN_RATIO}{P1_RATIO} =
\dfrac{\text{ plural(GIVEN_MOL, "mole")}}{x}
x = \text{ P1_MOL plural_form(MOLE, P1_MOL)} of P1 produced.