Question

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

Answer 1

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 2

Answer : The volume of
`CO_2` produced will be, 48 liters.

Explanation : Given,

Volume of
`O_2` = 60 L

First we have to calculate the moles of
`O_2`.

The given balanced chemical reaction is,

`2C_2H_2+5O_2\rightarrow 2H_2O+4CO_2`

From the balanced chemical reaction, we conclude that

As, 22.4 L volume of
`O_2` present in 1 mole of
`O_2`

So, 60 L volume of
`O_2` present in
`(60)/(22.4)=2.68` mole of
`O_2`

Now we have to calculate the moles of
`CO_2`.

From the reaction we conclude that,

As, 5 moles of
`O_2` react to give 4 moles of
`CO_2`

So, 2.68 moles of
`O_2` react to give
`(4)/(5)* 2.68=2.144` moles of
`CO_2`

Now we have to calculate the volume of
`CO_2`.

At STP,

As, 1 mole of
`CO_2` contains 22.4 L volume of
`CO_2`

So, 2.144 mole of
`CO_2` contains
`2.144* 22.4=48.0L` volume of
`CO_2`

Therefore, the volume of
`CO_2` produced will be, 48 liters.