# Limiting reagent problem? For the reaction C_2H_6+O_2 to CO_2+H_2O

## (1) Balance (2) if $\text{270 g}$ ${\text{C"_2"H}}_{6}$ and $\text{300 g}$ ${\text{O}}_{2}$ reacted, how many moles of ${\text{CO}}_{2}$ were produced? (3) How many grams excess leftover?

Feb 16, 2016

see below

#### Explanation:

Balanced Eqn
$2 {C}_{2} {H}_{6} + 7 {O}_{2} = 4 C {O}_{2} + 6 {H}_{2} O$
By the Balanced eqn
60g ethane requires 7x32= 224g oxygen
here ethane is in excess.oxygen will be fully consumed
hence
300g oxygen will consume $60 \cdot \frac{300}{224} = 80.36 g$ ethane
leaving (270-80.36)= 189.64 g ethane.
By the Balanced eqn
60g ethane produces 4x44 g CO2
hence amount of CO2 produced =$4 \cdot 44 \cdot \frac{80.36}{60} = 235.72 g$

and its no. of moles will be $\frac{235.72}{44}$ =5.36 where 44 is the molar mass of Carbon dioxide

Feb 18, 2016

Balanced equation is ${C}_{2} {H}_{6} + \frac{7}{2} {O}_{2} \to 2 C {O}_{2} + 3 {H}_{2} O$

#### Explanation:

Balanced equation is ${C}_{2} {H}_{6} + \frac{7}{2} {O}_{2} \to 2 C {O}_{2} + 3 {H}_{2} O$

Alternatively, this can be written as $2 {C}_{2} {H}_{6} + 7 {O}_{2} \to 4 C {O}_{2} + 6 {H}_{2} O$

Find the number of moles of each reactant:

For ${C}_{2} {H}_{6}$, with molar mass $30$ $g m o {l}^{-} 1$:

$n = \frac{m}{M} = \frac{270}{30} = 9$ $m o l$

For ${O}_{2}$, with molar mass $32$ $g m o {l}^{-} 1$:

$n = \frac{m}{M} = \frac{300}{32} = 9.375$ $m o l$

It might look as though we have an excess of ${O}_{2}$, but remember that each $1$ $m o l$ of ${C}_{2} {H}_{6}$ required $\frac{7}{2}$ $m o l$ of ${O}_{2}$, so in fact we have less oxygen than we need. Some ${C}_{2} {H}_{6}$ will remain unburned at the end of the reaction.

Divide the $9.375$ $m o l$ by $\frac{7}{2}$ to discover that $2.68$ $m o l$ of ${C}_{2} {H}_{6}$ will react. Each mole of ${C}_{2} {H}_{6}$ produces $2$ $m o l$ of $C {O}_{2}$, so $5.36$ $m o l$ of $C {O}_{2}$ will be produced.

The remaining amount of ${C}_{2} {H}_{6}$ will be $9 - 2.68 = 6.32$ $m o l$. To find the mass of excess ${C}_{2} {H}_{6}$:

$m = n M = 6.32 \cdot 30 = 189.6$ $g$