The theoretical yield of carbon dioxide would be 333.7 g while the percent yield would be 93.5%
From the equation of the reaction: 2C2H6(g) + 702(g) → 4CO2(g) + 6H2O
Mole ratio of C2H6 and CO2 = 2:4 or 1:2
Mole of 114 g C2H6 = 114/30.07
= 3.79 moles
Equivalent mole of CO2 = 3.79 x 2
= 7.58 moles
Theoretical yield of CO2 = 7.58 x 44.01
= 333.7 g
If 312 g is actually produced;
Percent yield = yield/theoretical yield x 100%
= 312/333.7 x 100%
= 93.5%
More on percent yield can be found here: https://brainly.com/question/2506978?referrer=searchResults
This question is providing the mass of ethane that is burnt in excess oxygen as 114 g. Thus, the theoretical and percent yields are required given 312 g of carbon dioxide is actually produced. At the end, the results turns out to be 334 g and 93.4 %.
In chemistry, we can analyze combustion reactions as those between a fuel, usually a hydrocarbon, and oxygen to produce carbon dioxide and water. Therefore, this problem's combustion reaction is:
[tex]2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O (g)[/tex]
Thus, we can calculate the mass of carbon dioxide product via stoichiometry.
In chemistry, we use stoichiometry as a tool to perform mole-mass relationships based on given balanced chemical equations. In such a way, since we are given 114 g of ethane, one can calculate the produced carbon dioxide with the following stoichiometric setup:
[tex]114gC_2H_6*\frac{1molC_2H_6}{30.07gC_2H_6}*\frac{4molCO_2}{2molC_2H_6}* \frac{44.01gCO_2}{1molCO_2}[/tex]
Where 30.07 is the molar mass of ethane, 4:2 the mole ratio of carbon dioxide to ethane and 44.01 the molar mass of carbon dioxide. Thereby, we obtain the following theoretical yield:
[tex]334gCO_2[/tex]
Finally, we calculate the percent yield with:
[tex]Y=\frac{actual\ yield}{theoretical\ yield} *100\%\\\\Y=\frac{312g}{334g}*100\%\\ \\Y=93.4\%[/tex]
Learn more about stoichiometry: brainly.com/question/9743981
