# Question bd124

Feb 4, 2018

$57.4 g$ $C {O}_{2}$

#### Explanation:

C_2H_5OH + 3O_2 → 2CO_2+ 3H_2O

1 mole of ${C}_{2} {H}_{5} O H$ produces 2 moles of $C {O}_{2}$

$\implies 46 g$ ${C}_{2} {H}_{5} O H$ produces $88 g$ $C {O}_{2}$

$\implies 1 g$ ${C}_{2} {H}_{5} O H$ produces $\frac{88}{46} g$ $C {O}_{2}$

$\implies 30 g$ ${C}_{2} {H}_{5} O H$ produces $\frac{88}{46} \cdot 30 g$ $C {O}_{2}$

$\implies 30 g$ ${C}_{2} {H}_{5} O H$ produces $57.4 g$ $C {O}_{2}$

Feb 4, 2018

Since ${\text{O}}_{2}$ is used for burning, it's usually in excess. The fuel (here ethanol) is the limiting reactant and so the yield of carbon dioxide will only depend on that.

#### Explanation:

$\text{Moles of ethanol" = "Mass of ethanol"/"Molar mass of ethanol}$

= ("30.0 g")/("46.07 g/mol") = "0.6512 mols"#

Since all of the $\text{C}$ in ethanol converts to ${\text{CO}}_{2}$, we can use the stoichiometric relationship from the balanced equation to relate the moles of ethanol to moles of ${\text{CO}}_{2}$.

${\text{1 mol ethanol " -> " 2 moles CO}}_{2}$

So, the number of moles of ${\text{CO}}_{2}$ produced by $0.65$ moles ethanol $= \text{0.6512 mol" *2 = "1.3024 moles}$

${\text{Mass of CO"_2 = "Moles CO"_2 * "Molar mass of CO}}_{2}$

$= \text{1.3024 moles" * "44.0 g/mol" = "57.3 g}$