Question

the chemical formula of petrol is C12H26.if we want to burn completly one litre in air.find out mass ,moles and volume of oxygen.volume of air (hint:O2 21% of air mixture).how much CO2 is produce by mass,volume.soppose 1 litre=1000g

Answer 1

Question#47243 – Chemistry – Organic Chemistry

Question:

The chemical formula of petrol is $C_{12}H_{26}$. If we want to burn completely 1 liter in air, find out mass of oxygen, moles of oxygen, volume of oxygen, volume of air (hint: $O_2$ 21% of air mixture). How much $CO_2$ is produced by mass, volume. Suppose 1 liter = 1000g.

Answer:

The equation of reaction:

$$2 C_{12}H_{26} + 37 O_2 \rightarrow 24 CO_2 + 26 H_2O$$

According to this equation, to burn 2 liters of hydrocarbon $C_{12}H_{26}$, 37 liters of oxygen $O_2$ are required. If we have 1 liter of hydrocarbon $C_{12}H_{26}$, than $V(O_2) = 37/2 = 18.5$ liters of oxygen $O_2$ are necessary.

The amount of moles of $O_2$ can be estimated:

$n(O_2) = \frac{V(O_2)}{V_m}$, where $V_m = 22.4 \, \text{L/mol}$ is the volume of one mole of gaseous compound at STP.

Therefore, the amount of moles of oxygen $O_2$:

$$n(O_2) = \frac{V(O_2)}{V_m} = \frac{18.5 \, \text{L}}{22.4 \, \text{L/mol}} = 0.826 \, \text{mol}$$

The mass of oxygen $O_2$ is defined as:

$m(O_2) = n(O_2) \times M(O_2) = 0.826 \, \text{mol} \times 31.999 \, \text{g/mol} = 26.4 \, \text{g}$, where $M(O_2)$ is the mass of one mole of oxygen $O_2$.

The volume of air required is greater than the volume of Oxygen:

$$V(\text{air}) = \frac{100\% \times V(O_2)}{21\%} = \frac{100\% \times 18.5 \, \text{L}}{21\%} = 88 \, \text{L}$$

The volume of $CO_2$ produced is twelve times greater than the volume of hydrocarbon $C_{12}H_{26}$ burnt, $V(CO_2) = 1L \times 12 = 12 \, \text{L}$.

The mass of $CO_2$:

$$m(CO_2) = n(CO_2) \times M(CO_2) = \frac{V(CO_2)}{V_m} \times M(CO_2) = \frac{12 \, \text{L}}{22.4 \, \text{L/mol}} \times 44 \, \text{g/mol} = 23.6 \, \text{g}$$