Question

find volume of C3H8 which will react to 30L of O2​

Answer 1

First of all, write down the balanced combustion reaction equation !

$\rm C_{3}H_{8}+5O_{2}\rightarrow 3CO_{2}+4H_{2}O +\Delta$

As per the Balanced Equation, we can say:

• 22.4 L of propane $(C_{3}H_{8})$ will react with 5 × 22.4 = 112 L of Oxygen.

Now, amount of propane needed for 30L of Oxygen is :

$= \dfrac{22.4}{112} \times 30$

$= 0.2 \times 30$

$= 6 \: litres$

So, 6 litres of propane is needed.

Answer 2

Explanation:

First of all, write down the balanced combustion reaction equation !

C

3

H

8

+5O

2

→3CO

2

+4H

2

O+Δ

As per the Balanced Equation, we can say:

22.4 L of propane (C_{3}H_{8})(C

3

H

8

) will react with 5 × 22.4 = 112 L of Oxygen.

Now, amount of propane needed for 30L of Oxygen is :

= \dfrac{22.4}{112} \times 30=

112

22.4

×30

= 0.2 \times 30=0.2×30

= 6 \: litres=6litres

So, 6 litres of propane is needed