quiz - gas laws\nname: ebtesam\nquestions\nuse the following information to answer the next question.\nthe…

quiz - gas laws\nname: ebtesam\nquestions\nuse the following information to answer the next question.\nthe unbalanced reaction equation for the combustion of propane gas is given as follows:\n__c₃h₈(g) + __o₂(g) → __co₂(g) + __h₂o(g)\n8. if 27l of propane gas is burned with excess oxygen, the expected volume of carbon dioxide produced will be ____ l.\nyour answer

quiz - gas laws\nname: ebtesam\nquestions\nuse the following information to answer the next question.\nthe unbalanced reaction equation for the combustion of propane gas is given as follows:\n__c₃h₈(g) + __o₂(g) → __co₂(g) + __h₂o(g)\n8. if 27l of propane gas is burned with excess oxygen, the expected volume of carbon dioxide produced will be ____ l.\nyour answer

Answer

Explanation:

Step1: Balance the chemical equation

$C_3H_8(g)+5O_2(g)\rightarrow3CO_2(g) + 4H_2O(g)$

Step2: Use the mole - volume relationship

At the same temperature and pressure, the volume ratio of gases is equal to the mole ratio. The ratio of propane ($C_3H_8$) to carbon dioxide ($CO_2$) is 1:3.

Step3: Calculate the volume of $CO_2$

If $V_{C_3H_8}=27L$, and $\frac{V_{C_3H_8}}{V_{CO_2}}=\frac{1}{3}$, then $V_{CO_2}=3\times V_{C_3H_8}$. So $V_{CO_2}=3\times27L = 81L$.

Answer:

81