# Limiting reactants and masses

## In a nutshell

The limiting reactant is the reactant that is used up first and therefore causes the reaction to finish. The amount of product can be calculated using the mass of the limiting reactant. Equations are able to be balanced using the masses of the reactants and products.

## Limiting reactants

### Definition

The limiting reactant is the reactant that is used up first in a reaction, leaving the other reactant(s) in excess. The limiting reactant is responsible for product formation, so the amount of product formed is directly proportional to the amount of the limiting reactant there is in the reaction.

To find the limiting reactant, a ratio between the number of moles must be found and then compared against the ratio between substances in the balanced equation. Once identified, the amount of product formed can be found.

#### PROCEDURE

1. | Find the number of moles for each reactant using their masses. $Moles=\frac{Mass}{M_r}$ |

2. | Divide each number of moles by the smallest number of moles to find a ratio. |

3. | Compare the ratio with the balanced equation. If one reactant has a smaller ratio compared to the equation, it is the limiting reactant. |

4. | Use the moles of the limiting reactant to find the number of moles for the product. |

5. | Find the mass of the product using the moles. $Mass=Moles \times M_r$ |

##### Example

*A reaction occurs between *$48.75 \space g$* of *$Zn$* and *$8 \space g$* of *$O_2$*. The balanced equation for this reaction is:*

$2Zn+O_2 \rightarrow 2ZnO$

*What is the limiting reactant? How much *$ZnO$* product is formed?*

*First calculate the number of moles of *$Zn$* and** *$O_2$*:*

$Moles(Zn)=\frac{48.75}{65}=0.75 \space moles\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,Moles(O_2)=\frac{8}{32}=0.25 \space moles$

*Divide each number of moles by the smallest mole value (*$0.25 \space moles$*) to find a ratio:*

$Zn=\frac{0.75}{0.25}=3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,O_2=\frac{0.25}{0.25}=1$

*Compare the ratio with the balanced equation:*

$Zn:O_2=3:1$ Balanced equation: $Zn:O_2=2:1$

*The *$O_2$* has a smaller ratio. Therefore *$O_2$* is the limiting reactant.*

*Use the moles of *$O_2$* to find the moles of the product (*$ZnO$*), based on the ratios in the balanced equation:*

$Moles(ZnO)=0.25 \times2=0.5 \space moles$

*Calculate the mass of *$ZnO$* using the moles:*

$Mass(ZnO)=0.5 \times 81=40.5 \space g$

*Therefore, the mass of the *$ZnO$* product is *$\underline{40.5\ g}$*. *

## Balancing equations using masses

Equations can be balanced using the masses of reactant and products in a reaction.

#### procedure

1. | Work out any unknown masses. If only one product is formed, the mass is the total mass of all reactants. |

2. | Find the number of moles of all reactants and products using their masses. $Moles=\frac{Mass}{M_r}$ |

3. | Divide each number of moles by the smallest number of moles to find a ratio. |

4. | Use the ratio to write the balanced equation. |

##### Example

$8.32\ g$* of an unknown metal (A) is burnt in *$3.84\ g$* of oxygen to produce a metal oxide. Write a balanced equation for this reaction. *

$A_r (A)= 52$ *and* $M_r (unknown \ metal\ oxide)=152$

*Work out the mass of unknown metal oxide produced: *

$8.32\ g+3.84\ g=12.16\ g$

*Work out the number of moles of the unknown metal A:*

$Moles=\frac{Mass}{M_r}$

$\frac{8.32}{52}=0.16\ mol$

*Work out the number of moles of oxygen:*

*$M_r=16+16=32$*

**

$\frac{3.84}{32}=0.12\ mol$

*Work out the number of moles of unknown metal oxide:*

$\frac{12.16}{152}=0.08\ mol$

*Divide each number of moles by the smallest number of moles to find a ratio:*

$A: \frac{0.16}{0.08}=2$ | $Oxygen: \frac{0.12}{0.08}=1.5$ | $Unknown\ metal\ oxide: \frac{0.08}{0.08}=1$ |

*Convert the ratio into whole numbers:*

$2:1.5:1=4:3:2$

*So, now you have the equation *$4A + 3O_2 \rightarrow 2\ (Unknown\ metal\ oxide)$*. The formula of the unknown metal oxide must be $A_2O_3$ in order for the equation to balance. *

*Therefore, the final balanced equation is $4A+3O_2\rightarrow 2A_2O_3$. *