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In the following combustion reaction of acetylene (C2H2), how many liters of CO2 will be produced if 60 liters of O2 is used, given that both gases are at STP? 2C2H2+5O2 2H2O+4CO2 The volume of one mole of gas at STP is 22.4 liters. 48 liters 0.02 liters 240 liters 300 liters NextReset

Respuesta :

V(O₂)=60 L

V(O₂)/5=V(CO₂)/4

V(CO₂)=4V(O₂)/5

V(CO₂)=4*60/5=48 L

48 liters

Answer : The volume of [tex]CO_2[/tex] produced will be, 48 liters.

Explanation : Given,

Volume of [tex]O_2[/tex] = 60 L

First we have to calculate the moles of [tex]O_2[/tex].

The given balanced chemical reaction is,

[tex]2C_2H_2+5O_2\rightarrow 2H_2O+4CO_2[/tex]

From the balanced chemical reaction, we conclude that

As, 22.4 L volume of [tex]O_2[/tex] present in 1 mole of [tex]O_2[/tex]

So, 60 L volume of [tex]O_2[/tex] present in [tex]\frac{60}{22.4}=2.68[/tex] mole of [tex]O_2[/tex]

Now we have to calculate the moles of [tex]CO_2[/tex].

From the reaction we conclude that,

As, 5 moles of [tex]O_2[/tex] react to give 4 moles of [tex]CO_2[/tex]

So, 2.68 moles of [tex]O_2[/tex] react to give [tex]\frac{4}{5}\times 2.68=2.144[/tex] moles of [tex]CO_2[/tex]

Now we have to calculate the volume of [tex]CO_2[/tex].

At STP,

As, 1 mole of [tex]CO_2[/tex] contains 22.4 L volume of [tex]CO_2[/tex]

So, 2.144 mole of [tex]CO_2[/tex] contains [tex]2.144\times 22.4=48.0L[/tex] volume of [tex]CO_2[/tex]

Therefore, the volume of [tex]CO_2[/tex] produced will be, 48 liters.