Question

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

In the equation, if 14 L of ethane (C2H6) and 14 L of oxygen (O2) combined and burned to completion, which gas will be leftover after the reaction, and what is the volume of that gas remaining

Answer 1

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

$14LO_2(\frac{2LC_2H_6}{7LO_2})$

= $4LC_2H_6$

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.