Question

what volume of carbon dioxide could be produced at 10.0 degrees celsius and 96.0 kpa, if 650g of ethanol (c₂h₅oh₍ₗ₎) is burned completely to produce carbon dioxide and water vapour? (show all steps including balanced equation) -(5 marks)

Explanation:

Step1: Write balanced chemical equation

$C_2H_5OH(l)+3O_2(g)
ightarrow2CO_2(g) + 3H_2O(g)$

Step2: Calculate moles of ethanol

The molar mass of $C_2H_5OH$ is $M=(2\times12.01 + 6\times1.01+16.00)\ g/mol = 46.08\ g/mol$.
The number of moles of ethanol, $n_{C_2H_5OH}=\frac{m}{M}=\frac{650\ g}{46.08\ g/mol}\approx14.1\ mol$

Step3: Determine moles of $CO_2$

From the balanced equation, the mole - ratio of $C_2H_5OH$ to $CO_2$ is $1:2$. So, $n_{CO_2}=2\times n_{C_2H_5OH}=2\times14.1\ mol = 28.2\ mol$

Step4: Use ideal gas law

The ideal gas law is $PV = nRT$. First, convert the temperature to Kelvin: $T=(10.0 + 273.15)\ K=283.15\ K$ and $P = 96.0\ kPa=96000\ Pa$. $R = 8.314\ J/(mol\cdot K)$.
We want to find $V$, so $V=\frac{nRT}{P}$.
Substitute the values: $V=\frac{28.2\ mol\times8.314\ J/(mol\cdot K)\times283.15\ K}{96000\ Pa}$
$V=\frac{28.2\times8.314\times283.15}{96000}\ m^3$
$V\approx0.694\ m^3$
Since $1\ m^3 = 1000\ L$, $V = 694\ L$

Answer:

$694\ L$