Question

combustion reaction 1/2
consider equation no. 9.

1. how many grams of co₂ are produced when 5.21 grams of (k) is reacted with excess oxygen? input the answer here.

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

1. ch₄ + o₂ → co₂ + h₂o

(a) 16.00 g/mol

1. c₂h₆ + o₂ → co₂ + h₂o

(b) 30.00 g/mol

1. c₃h₈ + o₂ → co₂ + h₂o

(c) 44.00 g/mol

1. c₄h₁₀ + o₂ → co₂ + h₂o

(d) 58.00 g/mol

1. c₅h₁₂ + o₂ → co₂ + h₂o

(e) 72.00 g/mol

1. c₆h₁₄ + o₂ → co₂ + h₂o

(f) 86.00 g/mol

1. c₇h₁₆ + o₂ → co₂ + h₂o

(g) 100.00 g/mol

1. c₈h₁₈ + o₂ → co₂ + h₂o

(h) 114.00 g/mol

1. c₉h₂₀ + o₂ → co₂ + h₂o

(k) 128.00 g/mol

1. c₁₀h₂₂ + o₂ → co₂ + h₂o

(m) 142.00 g/mol

1. c₇h₁₂ + o₂ → co₂ + h₂o

(p) 96.00 g/mol
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: Write balanced chemical equation

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

Step2: Calculate moles of $C_9H_{20}$

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

Step3: Determine mole - ratio

From the balanced equation, the mole - ratio of $C_9H_{20}$ to $CO_2$ is $1:9$. So, for every 1 mole of $C_9H_{20}$ reacted, 9 moles of $CO_2$ are produced.

Step4: Calculate moles of $CO_2$

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 $n_{CO_2}=9\times0.0407\ mol = 0.3663\ mol$.

Step5: Calculate mass of $CO_2$

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, $m_{CO_2}=0.3663\ mol\times44.00\ g/mol = 16.12\ g$.

Answer:

$16.12\ g$