Answer :
To solve the given problem, we need to analyze the reaction stoichiometry and how the given masses of [tex]\( O_2 \)[/tex] and [tex]\( H_2 \)[/tex] translate to volumes of [tex]\( H_2O \)[/tex] produced at Standard Temperature and Pressure (STP). Here, one mole of any gas at STP occupies 22.4 liters.
First, let's write down the balanced chemical equation:
[tex]\[ 2 H_2(g) + O_2(g) \longrightarrow 2 H_2O(g) \][/tex]
Step 1: Determine the molar masses.
- The molar mass of [tex]\( H_2 \)[/tex] is [tex]\( 2.00 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( O_2 \)[/tex] is [tex]\( 32.00 \, \text{g/mol} \)[/tex].
Step 2: Calculate the moles of the given masses.
Option 1: 32.00 g of [tex]\( O_2 \)[/tex] react with an excess of [tex]\( H_2 \)[/tex] to produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] of [tex]\( H_2O \)[/tex].
- Moles of [tex]\( O_2 \)[/tex] in 32.00 g:
[tex]\[ \text{moles of } O_2 = \frac{32.00 \, \text{g}}{32.00 \, \text{g/mol}} = 1 \, \text{mol} \][/tex]
- According to the stoichiometry of the balanced equation, 1 mole of [tex]\( O_2 \)[/tex] produces 2 moles of [tex]\( H_2O \)[/tex].
- Volume of [tex]\( H_2O \)[/tex] produced (at STP):
[tex]\[ \text{Volume of } H_2O = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L} \][/tex]
So, this option states correctly that 32.00 g of [tex]\( O_2 \)[/tex] produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] or 44.8 liters of [tex]\( H_2O \)[/tex].
Option 2: 2.00 g of [tex]\( H_2 \)[/tex] react with an excess of [tex]\( O_2 \)[/tex] to produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] of [tex]\( H_2O \)[/tex].
- Moles of [tex]\( H_2 \)[/tex] in 2.00 g:
[tex]\[ \text{moles of } H_2 = \frac{2.00 \, \text{g}}{2.00 \, \text{g/mol}} = 1 \, \text{mol} \][/tex]
- According to the stoichiometry of the balanced equation, 1 mole of [tex]\( H_2 \)[/tex] produces 1 mole of [tex]\( H_2O \)[/tex]. However, considering there are 2 moles of [tex]\( H_2 \)[/tex] reacting and producing 2 moles of [tex]\( H_2O \)[/tex], the volume here would be:
[tex]\[ \text{Volume of } H_2O = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L} \][/tex]
This option is identical to the first in terms of end results with respect to volume produced.
Option 3: 32.00 g of [tex]\( O_2 \)[/tex] react with an excess of [tex]\( H_2 \)[/tex] to produce 22.4 L of [tex]\( H_2O \)[/tex].
We previously calculated that 32.00 g of [tex]\( O_2 \)[/tex] produces 44.8 L of [tex]\( H_2O \)[/tex], not 22.4 L.
Option 4: 2.00 g of [tex]\( H_2 \)[/tex] react with an excess of [tex]\( O_2 \)[/tex] to produce 22.4 L of [tex]\( H_2O \)[/tex].
- This volume corresponds to just 1 mole of [tex]\( H_2O \)[/tex] produced, but we have 2 moles of [tex]\( H_2O \)[/tex] being produced from 2.00 g of [tex]\( H_2 \)[/tex].
Therefore, the correct mass-volume relationship is represented by Option 1:
[tex]\[ \boxed{32.00 \, \text{g of } O_2 \text{ react with an excess of } H_2 \text{ to produce } (2 \times 22.4) \, \text{L of } H_2O} \][/tex]
First, let's write down the balanced chemical equation:
[tex]\[ 2 H_2(g) + O_2(g) \longrightarrow 2 H_2O(g) \][/tex]
Step 1: Determine the molar masses.
- The molar mass of [tex]\( H_2 \)[/tex] is [tex]\( 2.00 \, \text{g/mol} \)[/tex].
- The molar mass of [tex]\( O_2 \)[/tex] is [tex]\( 32.00 \, \text{g/mol} \)[/tex].
Step 2: Calculate the moles of the given masses.
Option 1: 32.00 g of [tex]\( O_2 \)[/tex] react with an excess of [tex]\( H_2 \)[/tex] to produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] of [tex]\( H_2O \)[/tex].
- Moles of [tex]\( O_2 \)[/tex] in 32.00 g:
[tex]\[ \text{moles of } O_2 = \frac{32.00 \, \text{g}}{32.00 \, \text{g/mol}} = 1 \, \text{mol} \][/tex]
- According to the stoichiometry of the balanced equation, 1 mole of [tex]\( O_2 \)[/tex] produces 2 moles of [tex]\( H_2O \)[/tex].
- Volume of [tex]\( H_2O \)[/tex] produced (at STP):
[tex]\[ \text{Volume of } H_2O = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L} \][/tex]
So, this option states correctly that 32.00 g of [tex]\( O_2 \)[/tex] produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] or 44.8 liters of [tex]\( H_2O \)[/tex].
Option 2: 2.00 g of [tex]\( H_2 \)[/tex] react with an excess of [tex]\( O_2 \)[/tex] to produce [tex]\( 2 \times 22.4 \, \text{L} \)[/tex] of [tex]\( H_2O \)[/tex].
- Moles of [tex]\( H_2 \)[/tex] in 2.00 g:
[tex]\[ \text{moles of } H_2 = \frac{2.00 \, \text{g}}{2.00 \, \text{g/mol}} = 1 \, \text{mol} \][/tex]
- According to the stoichiometry of the balanced equation, 1 mole of [tex]\( H_2 \)[/tex] produces 1 mole of [tex]\( H_2O \)[/tex]. However, considering there are 2 moles of [tex]\( H_2 \)[/tex] reacting and producing 2 moles of [tex]\( H_2O \)[/tex], the volume here would be:
[tex]\[ \text{Volume of } H_2O = 2 \, \text{mol} \times 22.4 \, \text{L/mol} = 44.8 \, \text{L} \][/tex]
This option is identical to the first in terms of end results with respect to volume produced.
Option 3: 32.00 g of [tex]\( O_2 \)[/tex] react with an excess of [tex]\( H_2 \)[/tex] to produce 22.4 L of [tex]\( H_2O \)[/tex].
We previously calculated that 32.00 g of [tex]\( O_2 \)[/tex] produces 44.8 L of [tex]\( H_2O \)[/tex], not 22.4 L.
Option 4: 2.00 g of [tex]\( H_2 \)[/tex] react with an excess of [tex]\( O_2 \)[/tex] to produce 22.4 L of [tex]\( H_2O \)[/tex].
- This volume corresponds to just 1 mole of [tex]\( H_2O \)[/tex] produced, but we have 2 moles of [tex]\( H_2O \)[/tex] being produced from 2.00 g of [tex]\( H_2 \)[/tex].
Therefore, the correct mass-volume relationship is represented by Option 1:
[tex]\[ \boxed{32.00 \, \text{g of } O_2 \text{ react with an excess of } H_2 \text{ to produce } (2 \times 22.4) \, \text{L of } H_2O} \][/tex]
