Question

part b when carbon is burned in air, it reacts with oxygen to form carbon dioxide. when 26.4 g of carbon were burned in the presence of 86.8 g of oxygen, 16.4 g of oxygen remained unreacted. what mass of carbon dioxide was produced? express your answer to one decimal place and include the appropriate units. view available hint(s)

Explanation:

Step1: Write the chemical equation

The chemical equation for carbon - oxygen reaction is $C + O_{2}
ightarrow CO_{2}$. The molar mass of $C$ is $M_{C}=12.01\ g/mol$, the molar mass of $O_{2}$ is $M_{O_{2}} = 32.00\ g/mol$, and the molar mass of $CO_{2}$ is $M_{CO_{2}}=44.01\ g/mol$.

Step2: Determine the limiting reactant

The number of moles of carbon, $n_{C}=\frac{m_{C}}{M_{C}}=\frac{26.4\ g}{12.01\ g/mol}\approx2.2\ mol$. The number of moles of oxygen, $n_{O_{2}}=\frac{m_{O_{2}}}{M_{O_{2}}}=\frac{86.8 - 16.4}{32.00\ g/mol}=\frac{70.4\ g}{32.00\ g/mol}=2.2\ mol$. From the balanced chemical equation, the mole - ratio of $C$ to $O_{2}$ is $1:1$. So, they react in a 1:1 ratio and there is no limiting reactant.

Step3: Calculate the mass of $CO_{2}$ produced

According to the mole - ratio in the chemical equation, the number of moles of $CO_{2}$ produced is equal to the number of moles of $C$ (or $O_{2}$) that react. $n_{CO_{2}} = 2.2\ mol$. The mass of $CO_{2}$ produced, $m_{CO_{2}}=n_{CO_{2}}\times M_{CO_{2}}=2.2\ mol\times44.01\ g/mol = 96.822\ g\approx96.8\ g$.

Answer:

96.8 g