combustion of hydrocarbons such as pentane (c₅h₁₂) produces carbon dioxide, a \greenhouse gas.\ greenhouse…

combustion of hydrocarbons such as pentane (c₅h₁₂) produces carbon dioxide, a \greenhouse gas.\ greenhouse gases in the earths atmosphere can trap the suns heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.\n1. write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water.\n2. suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 16.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.

combustion of hydrocarbons such as pentane (c₅h₁₂) produces carbon dioxide, a \greenhouse gas.\ greenhouse gases in the earths atmosphere can trap the suns heat, raising the average temperature of the earth. for this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.\n1. write a balanced chemical equation, including physical state symbols, for the combustion of liquid pentane into gaseous carbon dioxide and gaseous water.\n2. suppose 0.350 kg of pentane are burned in air at a pressure of exactly 1 atm and a temperature of 16.0 °c. calculate the volume of carbon dioxide gas that is produced. be sure your answer has the correct number of significant digits.

Answer

Explanation:

Step1: Write balanced chemical equation

The combustion of pentane ($C_5H_{12}$) in oxygen ($O_2$) produces carbon - dioxide ($CO_2$) and water ($H_2O$). The balanced chemical equation is $C_5H_{12}(l)+8O_2(g)\rightarrow5CO_2(g) + 6H_2O(g)$.

Step2: Calculate moles of pentane

The molar mass of $C_5H_{12}$ is $M=(5\times12.01 + 12\times1.01)\ g/mol=72.17\ g/mol$. The mass of pentane is $m = 0.350\ kg=350\ g$. The number of moles of pentane, $n_{C_5H_{12}}=\frac{m}{M}=\frac{350\ g}{72.17\ g/mol}\approx4.85\ mol$.

Step3: Determine moles of carbon - dioxide

From the balanced equation, the mole ratio of $C_5H_{12}$ to $CO_2$ is $1:5$. So, $n_{CO_2}=5\times n_{C_5H_{12}}=5\times4.85\ mol = 24.25\ mol$.

Step4: Use ideal gas law to find volume of carbon - dioxide

The ideal gas law is $PV = nRT$, where $P = 1\ atm$, $n=n_{CO_2}=24.25\ mol$, $R = 0.0821\ L\cdot atm/(mol\cdot K)$, and $T=(16.0 + 273.15)\ K=289.15\ K$. Rearranging for $V$, we get $V=\frac{nRT}{P}$. Substituting the values: $V=\frac{24.25\ mol\times0.0821\ L\cdot atm/(mol\cdot K)\times289.15\ K}{1\ atm}\approx572\ L$.

Answer:

572 L