Q77P - Chemistry: Molecular Nature Of Matter And Change

Question

When 0.100 mol of carbon is burned in a closed vessel with 8.00 g of oxygen, how many grams of carbon dioxide can form? Which reactant is in excess, and how many grams of it remain after the reaction?

Step-by-Step Solution

Answer

1. The mass of${}^{{\mathbf{CO}}_{\mathbf{2}}}$ is 4.401 g.
2. ${}^{{\mathbf{O}}_{\mathbf{2}}}$ is present in excess.
3. The mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$ left is 4.8 g.

Step 1: Balance the chemical equation

In a balanced chemical equation, the number of atoms on the reactant side should be the same as the number of atoms on the product side for a particular atom. For balancing the reaction, the atoms are multiplied by the stoichiometric coefficient.

So, the balanced chemical equation is:

$\mathbf{C}\left(s\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\mathbf{\to }{\mathbf{CO}}_{\mathbf{2}}\left(g\right)$

Step 2: Relation between mass and number of moles

The number of moles is calculated by the mass and Molar mass. The relationship between the number of moles, mass, and molar mass is given below.

$\mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{moles}\mathbf{=}\frac{\mathbf{mass}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}}$

Step 3: Calculate the number of moles of O 2

The mass of ${}^{{\mathbf{O}}_{\mathbf{2}}}$ = 8.00 g.

The molar mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$ = 31.996 g/mol

Thus, the number of moles${}^{{\mathbf{O}}_{\mathbf{2}}}$ is:

$$\begin{gathered} \mathbf{Number}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\frac{\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\frac{\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{g}}{\mathbf{31}\mathbf{.}\mathbf{996}\mathbf{g}\mathbf{/}\mathbf{mol}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{25}\mathbf{ }\mathbf{mol} \end{gathered}$$

Step 4: Determine the Limiting reagent of the reaction

In the reaction, 1 mol of${}^{\mathbf{C}}$ reacts with 1 mol of ${}^{{\mathbf{O}}_{\mathbf{2}}}$ Thus,

    $$\begin{gathered} \mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{molof}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}} \end{gathered}$$

Therefore,${}^{{\mathbf{O}}_{\mathbf{2}}}$ present in access in the reaction, hence $\mathit{C}$ is a limiting reagent.

Step 5: Calculate the number of moles of CO 2

In the given reaction, 1 mol of${}^{\mathbf{C}}$ forms 1 mol of ${}^{{\mathbf{CO}}_{\mathbf{2}}}$.

Thus, the number of moles of${}^{{\mathbf{CO}}_{\mathbf{2}}}$ is

$$\begin{gathered} \mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{1}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }\mathit{C}{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{molof}\mathbf{ }\mathbf{C}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{ }\mathbf{mol}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}} \end{gathered}$$

Step 6: Calculate the mass of CO 2

The molar mass of${}^{C{O}_2}$ = 44.01 g/mol

Thus, the number of moles of${}^{C{O}_2}$ is:

$$\begin{gathered} \mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}\mathbf{\times }\mathbf{molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{100}\mathbf{mol}\mathbf{\times }\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{44}\mathbf{.}\mathbf{01}\mathbf{ }\mathbf{g} \end{gathered}$$

Step 7: Calculate the mass of O 2 left

The molar mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$ = 31.998 g/mol

The number of moles${}^{{\mathbf{O}}_{\mathbf{2}}}$ left = 0.25 mol – 0.100 mol = 0.15 mol

Thus, the mass of${}^{{\mathbf{O}}_{\mathbf{2}}}$ left is:

$$\begin{gathered} \mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{ }\mathbf{left}\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{moles}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}}\mathbf{left}\mathbf{\times }\mathbf{molar}\mathbf{ }\mathbf{mass}\mathbf{ }\mathbf{of}\mathbf{ }{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{0}\mathbf{.}\mathbf{15}\mathbf{mol}\mathbf{\times }\mathbf{31}\mathbf{.}\mathbf{998}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{=}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{4}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{g} \end{gathered}$$