Question

combustion reaction 2/2
consider equation no. 9. how many grams of co₂ are produced when 5.21 grams of (k) is reacted with excess oxygen?

1. what percent yield is obtained if 10.2 grams of co₂ are collected after the reaction? input the numeric answer here.

hint: the equation is not balanced. table mm (g/mol): h₂o = 18.00, co₂ = 44.00.

1. ch₄ (a)16.00 g/mol + o₂ → co₂ + h₂o
2. c₂h₆ (b)30.00 g/mol + o₂ → co₂ + h₂o
3. c₃h₈ (c)44.00 g/mol + o₂ → co₂ + h₂o
4. c₄h₁₀ (d)58.00 g/mol + o₂ → co₂ + h₂o
5. c₅h₁₂ (e)72.00 g/mol + o₂ → co₂ + h₂o
6. c₆h₁₄ (f)86.00 g/mol + o₂ → co₂ + h₂o
7. c₇h₁₆ (g)100.00 g/mol + o₂ → co₂ + h₂o
8. c₈h₁₈ (h)114.00 g/mol + o₂ → co₂ + h₂o
9. c₉h₂₀ (k)128.00 g/mol + o₂ → co₂ + h₂o
10. c₁₀h₂₂ (m)142.00 g/mol + o₂ → co₂ + h₂o
11. c₇h₁₂ (p)96.00 g/mol + o₂ → co₂ + h₂o

this problem can be solved by dimensional analysis as the primary method, formula where permitted, or simple math. using ratio and proportion methods is not allowed. be sure to show the formula on homework and exams.
show work here as a file upload in the last question. be sure to report the units for all values when showing your work. keys will be posted for exemplary practice problems.
you must show work on the exams. practice showing the calculator answer and the sig figs answer.

Explanation:

Step1: Balance the combustion equation

The balanced equation for the combustion of $C_9H_{20}$ is $C_9H_{20}+14O_2
ightarrow9CO_2 + 10H_2O$.

Step2: Calculate the moles of $C_9H_{20}$

The molar - mass of $C_9H_{20}$ is $128.00\ g/mol$. The number of moles of $C_9H_{20}$, $n_{C_9H_{20}}=\frac{m_{C_9H_{20}}}{M_{C_9H_{20}}}$, where $m_{C_9H_{20}} = 5.21\ g$ and $M_{C_9H_{20}}=128.00\ g/mol$. So, $n_{C_9H_{20}}=\frac{5.21\ g}{128.00\ g/mol}=0.0407\ mol$.

Step3: Calculate the moles of $CO_2$ produced

From the balanced equation, the mole - ratio of $C_9H_{20}$ to $CO_2$ is $1:9$. So, the number of moles of $CO_2$, $n_{CO_2}=9\times n_{C_9H_{20}}$. Substituting the value of $n_{C_9H_{20}}$, we get $n_{CO_2}=9\times0.0407\ mol = 0.3663\ mol$.

Step4: Calculate the mass of $CO_2$ produced

The molar - mass of $CO_2$ is $44.00\ g/mol$. The mass of $CO_2$, $m_{CO_2}=n_{CO_2}\times M_{CO_2}$. So, $m_{CO_2}=0.3663\ mol\times44.00\ g/mol = 16.12\ g$.

Step5: Calculate the percent yield

The percent yield formula is $\text{Percent Yield}=\frac{\text{Actual Yield}}{\text{Theoretical Yield}}\times100\%$. The actual yield is $10.2\ g$ and the theoretical yield is $16.12\ g$. So, $\text{Percent Yield}=\frac{10.2\ g}{16.12\ g}\times100\% = 63.3\%$.

Answer:

1. $16.12\ g$
2. $63.3$