Question

use the following information to answer the next question. the unbalanced reaction equation for the combustion of propane gas is given as follows: \\_\\_\\_c3h8(g)+\\_\\_\\_o2(g)→\\_\\_\\_co2(g)+\\_\\_\\_h2o(g) 10. if 23l of propane gas is burned with excess oxygen, the expected volume of carbon dioxide produced will be \\_\\_\\_l your answer:

Explanation:

Step1: Balance the chemical equation

The balanced equation for the combustion of propane ($C_3H_8$) is $C_3H_8(g)+5O_2(g)
ightarrow3CO_2(g) + 4H_2O(g)$.

Step2: Use the mole - volume relationship

At the same temperature and pressure, the volume ratio of gases is equal to the mole ratio. From the balanced equation, the mole ratio of $C_3H_8$ to $CO_2$ is $1:3$. So the volume ratio of $C_3H_8$ to $CO_2$ is also $1:3$.

Step3: Calculate the volume of $CO_2$

If the volume of $C_3H_8$ is $V_{C_3H_8}=23L$, and $\frac{V_{C_3H_8}}{V_{CO_2}}=\frac{1}{3}$, then $V_{CO_2}=3\times V_{C_3H_8}$. Substituting the value of $V_{C_3H_8}$, we get $V_{CO_2}=3\times23L = 69L$.

Answer:

69L