Question

benzene (c₆h₆) is a liquid with a density of 0.874 g/cm³. it reacts with oxygen to give carbon dioxide (co₂) and water. calculate the mass of carbon dioxide that would form if 28.9 cm³ of benzene reacted completely with excess oxygen. mass = g
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Explanation:

Step1: Calculate the mass of benzene

Use the density - volume formula $m =
ho V$. Given $
ho=0.874\ g/cm^{3}$ and $V = 28.9\ cm^{3}$.
$m=
ho V=0.874\ g/cm^{3}\times28.9\ cm^{3}=25.2586\ g$

Step2: Write the balanced chemical equation for the combustion of benzene

$2C_{6}H_{6}+15O_{2}
ightarrow12CO_{2}+6H_{2}O$

Step3: Calculate the molar mass of benzene and carbon - dioxide

The molar mass of $C_{6}H_{6}$ is $M_{C_{6}H_{6}}=6\times12.01\ g/mol + 6\times1.01\ g/mol=72.06\ g/mol+6.06\ g/mol = 78.12\ g/mol$.
The molar mass of $CO_{2}$ is $M_{CO_{2}}=12.01\ g/mol + 2\times16.00\ g/mol=44.01\ g/mol$.

Step4: Calculate the number of moles of benzene

$n_{C_{6}H_{6}}=\frac{m_{C_{6}H_{6}}}{M_{C_{6}H_{6}}}=\frac{25.2586\ g}{78.12\ g/mol}=0.3233\ mol$

Step5: Calculate the number of moles of carbon - dioxide produced

From the balanced equation, the mole ratio of $C_{6}H_{6}$ to $CO_{2}$ is $2:12 = 1:6$. So, $n_{CO_{2}}=6\times n_{C_{6}H_{6}}=6\times0.3233\ mol = 1.94\ mol$

Step6: Calculate the mass of carbon - dioxide

$m_{CO_{2}}=n_{CO_{2}}\times M_{CO_{2}}=1.94\ mol\times44.01\ g/mol = 85.4\ g$

Answer:

$85.4$