$10\text{ g}$ of hydrogen and $64\text{ g}$ of oxygen were filled in a steel vessel and exploded. The amount of water produced in this reaction will be:

The balanced chemical equation for the formation of water is: $2\text{H}_2(g) + \text{O}_2(g) \rightarrow 2\text{H}_2\text{O}(g)$

First, we calculate the number of moles of each reactant: Molar mass of $\text{H}_2 = 2\text{ g/mol}$. Moles of $\text{H}_2 = \frac{10\text{ g}}{2\text{ g/mol}} = 5\text{ moles}$.

Molar mass of $\text{O}_2 = 32\text{ g/mol}$. Moles of $\text{O}_2 = \frac{64\text{ g}}{32\text{ g/mol}} = 2\text{ moles}$.

Now, we determine the limiting reagent. From the balanced equation, $2\text{ moles}$ of $\text{H}_2$ react with $1\text{ mole}$ of $\text{O}_2$. Thus, for $2\text{ moles}$ of $\text{O}_2$, the required moles of $\text{H}_2$ would be $2 \times 2 = 4\text{ moles}$. Since $5\text{ moles}$ of $\text{H}_2$ are present (which is more than the required $4\text{ moles}$), $\text{O}_2$ is the limiting reagent and $\text{H}_2$ is in excess.

The amount of product formed depends completely on the limiting reagent ($\text{O}_2$). According to the stoichiometry, $1\text{ mole}$ of $\text{O}_2$ produces $2\text{ moles}$ of $\text{H}_2\text{O}$. Therefore, $2\text{ moles}$ of $\text{O}_2$ will produce $2 \times 2 = 4\text{ moles}$ of $\text{H}_2\text{O}$.