A mixture of 0.0375 g of hydrogen and 0.0185 mol of oxygen in a closed container is sparked to initiate a reaction. How many grams of water can form? Which reactant is in excess, and how many grams of it remain after the reaction?

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:

$\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\left(g\right)\mathbf{+}{\mathbf{O}}_{\mathbf{2}}\left(g\right)\underset{\mathbf{}}{\mathbf{\to }}\mathbf{2}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\left(I\right)$

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.

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

The mass of ${\mathbf{H}}_{\mathbf{2}}$= 0.0375 g.

The molar mass of${\mathbf{H}}_{\mathbf{2}}$=2.02g/mol

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

$$\begin{gathered} \mathbf{Number}\mathbf{}\mathbf{of}\mathbf{}\mathbf{moles}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}}{\mathbf{Molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\frac{\mathbf{0}\mathbf{.}\mathbf{0375}\mathbf{}\mathbf{g}}{\mathbf{2}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{m}\mathbf{}\mathbf{ol}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{}\mathbf{}\mathbf{mol} \end{gathered}$$

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

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{\times }\frac{\mathbf{1}}{\mathbf{2}}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{O}}_{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{093}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{O}}_{\mathbf{2}} \end{gathered}$$

Therefore, ${\mathbf{O}}_{\mathbf{2}}$present in access in the reaction, hence ${\mathbf{H}}_{\mathbf{2}}$is alimiting reagent.

In the given reaction,2 mol of ${\mathbf{H}}_{\mathbf{2}}$forms 2 mol of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$

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

$$\begin{gathered} \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{2}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{1}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{}\mathbf{mol}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \end{gathered}$$

The molar mass of ${\mathbf{H}}_{\mathbf{2}}\mathbf{O}$= 18.02 g/mol

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

$$\begin{gathered} \mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{moles}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\times }\mathbf{molar}\mathbf{}\mathbf{mass}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{H}}_{\mathbf{2}}\mathbf{O} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{0186}\mathbf{}\mathbf{mol}\mathbf{\times }\mathbf{18}\mathbf{.}\mathbf{02}\mathbf{}\mathbf{g}\mathbf{/}\mathbf{mol} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{34}\mathbf{}\mathbf{g} \end{gathered}$$

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

The number of moles${\mathbf{O}}_{\mathbf{2}}$left = 0.0185 mol – 0.0093 mol = 0.0092 mol

Thus, themassof${\mathbf{O}}_{\mathbf{2}}$left is: