Question

1. balance the chemical equation below:

____c₂h₆ + __o₂ → __co₂ + ____h₂o
in an experiment conducted by a student, 3.50 g c₂h₆ burns in air with 2.7 l o₂ to produce 2.25 g co₂.
(a) calculate the theoretical yield or maximum amount in grams of co₂ produced in the reaction below hint: molar mass c₂h₆ =30.07 g/mol; molar mass co₂ = 44.01 g/mol; 1 mol any gas = 22.4 l of that gas
? g co₂ = 3.50 g c₂h₆
? g co₂ = 2.7 l o₂
(b) what is the percent yield of this reaction?

Explanation:

Step1: Balance the chemical equation

We use the method of balancing atoms on both sides. For the combustion of ethane ($C_2H_6$), the balanced equation is $2C_2H_6+7O_2
ightarrow4CO_2 + 6H_2O$.

Step2: Calculate moles of $C_2H_6$

The molar - mass of $C_2H_6$ is $M_{C_2H_6}=30.07\ g/mol$. The number of moles of $C_2H_6$, $n_{C_2H_6}=\frac{m_{C_2H_6}}{M_{C_2H_6}}=\frac{3.50\ g}{30.07\ g/mol}\approx0.1164\ mol$.
From the balanced equation, the mole - ratio of $C_2H_6$ to $CO_2$ is $2:4$ or $1:2$. So the moles of $CO_2$ produced from $C_2H_6$, $n_{CO_2\ from\ C_2H_6}=2\times n_{C_2H_6}=2\times0.1164\ mol = 0.2328\ mol$.
The mass of $CO_2$ from $C_2H_6$, $m_{CO_2\ from\ C_2H_6}=n_{CO_2\ from\ C_2H_6}\times M_{CO_2}=0.2328\ mol\times44.01\ g/mol\approx10.25\ g$.

Step3: Calculate moles of $O_2$

The volume of $O_2$ is $V_{O_2}=2.7\ L$. At standard conditions, $1\ mol$ of any gas is $22.4\ L$. So the number of moles of $O_2$, $n_{O_2}=\frac{V_{O_2}}{22.4\ L/mol}=\frac{2.7\ L}{22.4\ L/mol}\approx0.1205\ mol$.
From the balanced equation, the mole - ratio of $O_2$ to $CO_2$ is $7:4$. So the moles of $CO_2$ produced from $O_2$, $n_{CO_2\ from\ O_2}=\frac{4}{7}\times n_{O_2}=\frac{4}{7}\times0.1205\ mol\approx0.0689\ mol$.
The mass of $CO_2$ from $O_2$, $m_{CO_2\ from\ O_2}=n_{CO_2\ from\ O_2}\times M_{CO_2}=0.0689\ mol\times44.01\ g/mol\approx3.03\ g$.
The limiting reactant is $O_2$ (since it produces less $CO_2$), and the theoretical yield of $CO_2$ is $m_{theoretical}=3.03\ g$.

Step4: Calculate percent yield

The percent yield is given by the formula $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$.
The actual yield of $CO_2$ is $m_{actual}=2.25\ g$.
So the percent yield $=\frac{2.25\ g}{3.03\ g}\times100\%\approx74.3\%$.

Answer:

(a) The theoretical yield of $CO_2$ is $3.03\ g$.
(b) The percent yield of the reaction is $74.3\%$.