A mixture of \(0.0375 \mathrm{~g}\) of hydrogen and \(0.0185 \mathrm{~mol}\) 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 chemical reaction, a balanced chemical equation shows the relationship between the reactants and the products. For the reaction between hydrogen (\text{H}_2) and oxygen (\text{O}_2) to form water (\text{H}_2O), the balanced equation is: \[2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\] This equation tells us that 2 molecules of hydrogen react with 1 molecule of oxygen to produce 2 molecules of water. Make sure the number of atoms of each element is the same on both sides of the equation. This balance shows the conservation of mass. It's essential to always have a balanced chemical equation before performing any calculations.

Moles represent the amount of a substance. To find the number of moles, use the formula: \[ \text{Moles} = \frac{\text{Mass (g)}}{\text{Molar mass (g/mol)}} \] In our exercise, we need to calculate the moles of hydrogen and oxygen. The molar mass of hydrogen (\text{H}_2) is 2.016 g/mol. So for 0.0375 grams of hydrogen: \[ \text{Moles of } \text{H}_2 = \frac{0.0375 \text{ g}}{2.016 \text{ g/mol}} = 0.0186 \text{ mol} \] This tells us how many molecules of hydrogen are present. Remember, the molar mass is the weight of 1 mole of a substance.

Stoichiometry deals with the quantitative relationships in chemical reactions. The stoichiometric ratio comes from the coefficients of the balanced chemical equation. For \text{H}_2 and \text{O}_2, in the equation, the stoichiometric ratio is 2:1, meaning 2 moles of \text{H}_2 react with 1 mole of \text{O}_2. To determine which reactant runs out first (the limiting reactant), compare the available moles with the required stoichiometric ratio. In the exercise, we have 0.0186 mol \text{H}_2 and 0.0185 mol \text{O}_2. For 0.0186 mol \text{H}_2, the required \text{O}_2 is: \[ \frac{0.0186 \text{ mol } \text{H}_2}{2} = 0.0093 \text{ mol } \text{O}_2 \] Since 0.0185 mol \text{O}_2 is available, \text{H}_2 is the limiting reactant.

The excess reactant is the reactant that remains after the limiting reactant is consumed. In our case, oxygen (\text{O}_2) is in excess. To determine how much remains, calculate how much \text{O}_2 reacts: \[0.0186 \text{ mol } \text{H}_2 \times \frac{1 \text{ mol } \text{O}_2}{2 \text{ mol } \text{H}_2} = 0.0093 \text{ mol } \text{O}_2 \] Subtract this from the original amount: \[0.0185 \text{ mol } - 0.0093 \text{ mol } = 0.0092 \text{ mol } \text{O}_2 \] Convert this to grams: \[0.0092 \text{ mol } \text{O}_2 \times 32.00 \text{ g/mol} = 0.294 \text{ g} \text{O}_2 \] So, 0.294 grams of \text{O}_2 remains after the reaction. Understanding and calculating the excess reactant is crucial for industrial processes and cost efficiency.