Question follows after the explosion :Mole concept 3...

In the present case, $V \alpha n$ . As such all the volumes are measured under identical conditions of temperature and pressure. Hence the relation stoichiometry can be solved using volumes as : $C_n H_m (g) + \Big( n + \dfrac{m}{4} \Big) O_2 (g) \longrightarrow nCO_2 + \dfrac{m}{2} H_2 O (l)$ volume of $CO_2(g) + O_2(g)$ which remains unreacted = $25 mL$ $\Rightarrow$ Volume of $CO_2(g)$ produced = $10 mL$ as such $15 mL$ was pure oxygen Also $1 mL C_n H_m \text{produces n mL of } CO_2$ $\Rightarrow 5 mL C_n H_m \text{produces 5n mL of } CO_2 = 10 mL$ thus $\boxed{n = 2}$ Also $1mL C_n H_m \text{combines with } \Big( n + \dfrac{m}{4} \Big) mL \text{ of} O_2$ Hence equating the above for $5 mL$ we get the following equation $5 \Big( n + \dfrac{m}{4} \Big) = 15$ because $15 mL$ out of $30 mL$ remained unreacted. And so $\boxed{m = 4}$ And our gas is $\boxed{C_2 H_4}$

The gas $\ce{C_{n} H_{m}}$ is a hydrocarbon and combustion with $\ce{O2}$ gives $\ce{CO2}$ and $\ce{H2O}$ as follows:

$\ce{a C_{n} H_{m} (g) + b O2 (g) -> c CO2 (g) + d H2O (l)}$

At the same temperature and pressure, the mole of a gas is proportional to its volume. The volumes of the gases are as follows:

• $\ce{C_{n} H_{m}}$ used was $\text{5 mL}$
• The $\text{25 mL}$ mixed gases were $\ce{CO2}$ and $\ce{O2}$
• $\ce{KOH}$ reacted with all the $\ce{CO2}$ and reduced it to $\text{15 mL}$ of $\ce{O2}$
• Therefore, $\ce{CO2}$ produced was $25-15 = \text{10 mL}$
• And, $\ce{O2}$ consumed was $30-15=\text{15 mL}$

Therefore, we note that $a:b:c = 5:15:10 = 1:3:2$ and we have:

$\ce{C_{n} H_{m} (g) + 3 O2 (g) -> 2 CO2 (g) + d H2O (l)}$

• Balancing the $\ce{C}$ atoms of the reaction equation, we have $n = 2$
• Balancing the $\ce{O}$ atoms, we have $6 = 4 + d \space \Rightarrow d = 2$
• Balancing the $\ce{H}$ atoms, we have $m = 2d = 4$

Therefore, $n+m = 2 + 4 = \boxed{6}$