Question

Acetylene gas (c2h2) reacts with oxygen gas (o2) to produce carbon dioxide (co2) and water vapor (h2o). how many liters of c2h2 are required to produce 75.0 l of co2? l what volume of h2o is produced? l what volume of o2 is required? l when making the calculations, did you need to find the number of moles?

Answer 1

According to the balanced equation of the reaction:

2C2H2 + 5O2 → 4CO2 + 2H2O

So we can mention all as liters,

A) as we see that 2 liters of C2H2 react with 5 liters of oxygen to produce 4 liters of CO4 and 2 liters of H2O 

So, when we have 75L of CO2

and when we have 2 L of C2H2 reacts and gives 4 L of CO2

 2C2H2 → 4CO2 

∴ The volume of C2H2 required is:

                    = 75L / 2

                    = 37.5 L

B) and, when we have 75 L of CO2


and 4CO2 → 2H2O 

∴ the volume of H2O required is: 
                               
                            = 75 L /2
                            = 37.5 L 


C) and from the balanced equation and by  the same way:

when 5 liters O2  reacts to give 4 liters of CO2

and we have 75 L of CO2:

5 O2 → 4 CO2 

   ??  ←  75 L

∴ the volume of O2 required is:

                              = 75 *(5/4)

                              = 93.75 L



D) about the using of the number of moles the answer is: 

no, there is no need to find the number of moles as we called everything in the balanced equation by liters and use it as a liter unit to get the volume, without the need to get the number of moles.

Answer 2

Acetylene gas undergoes combustion under the presence of oxygen to yield, water vapour and carbon dioxide gas. The reaction generates some amount of heat and light as flames.

The volume of acetylene gas is 37.5 L

The volume of water is 37.5 L

The volume of oxygen is 93.75 L

No, moles are not needed

What are volume and moles?

The balanced chemical reaction is given as:

$\rm 2C_{2}H_{2} + 5O_{2} \rightarrow 4CO_{2} + 2H_{2}O$

2 litres of acetylene gas react with 5 litres of oxygen to yield 4 litres of carbon dioxide and 2 litres of water.

So when,

$\rm 2C_{2}H_{2} \rightarrow 4CO_{2}$

2 L of $\rm C_{2}H_{2}$ = 4 L of $\rm CO_{2}$

X litres of acetylene gas = 75 L of $\rm CO_{2}$

Solving X for acetylene:

$\begin{aligned}\text{The volume of} \;\rm 2C_{2}H_{2} &= \dfrac{75\;\rm L} { 2}\\\\&= 37.5\;\rm L\end{aligned}$

Hence, the volume of acetylene gas required is 37.5 L.

Similarly,

$\rm 4CO_{2} \rightarrow 2 H_{2}O$

4 L of $\rm CO_{2}$ = 2 L of $\rm H_{2}O$

75 L of $\rm CO_{2}$ = X liters of water

Solving X for water:

$\begin{aligned} \text{The volume of water} &= \dfrac{75\;\rm L} { 2}\\\\&= 37.5 \;\rm L\end{aligned}$

Hence, the volume of water required is 37.5 L.

Again for the oxygen we have,

$\rm 5 O_{2} \rightarrow 4 CO_{2}$

5 L of $\rm O_{2}$ = 4 L of $\rm CO_{2}$

X litres of oxygen = 75 L of $\rm CO_{2}$

Solving X for oxygen:

$\begin{aligned} \text{The volume of oxygen} &= \dfrac{75\;\rm L \times 5} { 4}\\\\&= 93.75\;\rm L\end{aligned}$

Hence, the volume of oxygen required is 93.75 L.

No, there is no need to calculate the number of moles as the reaction has the volume given in the litres and the volume required can be calculated without estimating the moles.

Therefore,

• The volume of acetylene gas is 37.5 L
• The volume of water is 37.5 L
• The volume of oxygen is 93.75 L
• No, moles are not needed.

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