Chapter 3: Problem 88
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?
Short Answer
Expert verified
0.335 g of water can form. Oxygen is in excess, and 0.294 g of it remains.
Step by step solution
01
- Write the Balanced Chemical Equation
The balanced chemical equation for the reaction between hydrogen (\text{H}_2) and oxygen (\text{O}_2) to form water (\text{H}_2\text{O}) is: \[2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O}\]
02
- Determine the Moles of Each Reactant
Calculate the moles of hydrogen using its molar mass (2.016 g/mol). \[ \text{Moles of } \text{H}_2 = \frac{0.0375 \text{ g}}{2.016 \text{ g/mol}} = 0.0186 \text{ mol} \]
03
- Identify the Limiting Reactant
From the equation, 2 moles of \text{H}_2 react with 1 mole of \text{O}_2. Compare the ratio of the available moles to the required stoichiometric ratio.Available moles: 0.0186 mol \text{H}_2 and 0.0185 mol \text{O}_2. Required ratio: \[ \frac{0.0186 \text{ mol } \text{H}_2}{2} = 0.0093 \text{ mol } \text{O}_2 \quad (\text{required})\]Since we have 0.0185 \text{ mol } \text{O}_2 available, \text{H}_2 is the limiting reactant.
04
- Calculate the Amount of Water Formed
Since we have 0.0186 mol of \text{H}_2 as the limiting reactant, it can completely react to form: \[0.0186 \text{ mol } \text{H}_2 \times \frac{2 \text{ moles } \text{H}_2\text{O}}{2 \text{ moles } \text{H}_2} = 0.0186 \text{ mol } \text{H}_2\text{O}\]
05
- Convert Moles of Water to Grams
The molar mass of water (\text{H}_2\text{O}) is 18.016 g/mol. Thus, the grams of water formed is: \[0.0186 \text{ mol } \text{H}_2\text{O} \times 18.016 \text{ g/mol} = 0.335 \text{ g } \text{H}_2\text{O} \]
06
- Determine the Excess Reactant and Its Remaining Amount
Since \text{H}_2 is the limiting reactant, \text{O}_2 is in excess. The amount of \text{O}_2 that reacts is: \[0.0186 \text{ mol } \text{H}_2 \times \frac{1 \text{ mol } \text{O}_2}{2 \text{ mol } \text{H}_2} = 0.0093 \text{ mol } \text{O}_2 \]The remaining \text{O}_2 is: \[0.0185 \text{ mol } - 0.0093 \text{ mol } = 0.0092 \text{ mol } \text{O}_2 \]Convert the remaining moles of \text{O}_2 to grams: \[0.0092 \text{ mol } \text{O}_2 \times 32.00 \text{ g/mol} = 0.294 \text{ g} \text{O}_2 \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
balanced chemical equation
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 calculation
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.
stoichiometric ratio
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.
excess 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.