Answer :
To solve this problem, we need to determine how much water is produced when hydrogen reacts with oxygen according to the balanced chemical equation:
[tex]\[ 2 \, \text{H}_2 + \text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} \][/tex]
Step 1: Calculate the moles of hydrogen ([tex]\( \text{H}_2 \)[/tex]) and oxygen ([tex]\( \text{O}_2 \)[/tex])
- Hydrogen ([tex]\( \text{H}_2 \)[/tex]):
Given mass of [tex]\( \text{H}_2 = 3.7 \)[/tex] grams
Molar mass of [tex]\( \text{H}_2 = 2.02 \)[/tex] g/mol
[tex]\[ \text{Moles of } \text{H}_2 = \frac{3.7 \, \text{g}}{2.02 \, \text{g/mol}} = 1.83 \, \text{moles} \][/tex]
- Oxygen ([tex]\( \text{O}_2 \)[/tex]):
Given mass of [tex]\( \text{O}_2 = 46.5 \)[/tex] grams
Molar mass of [tex]\( \text{O}_2 = 32 \)[/tex] g/mol
[tex]\[ \text{Moles of } \text{O}_2 = \frac{46.5 \, \text{g}}{32 \, \text{g/mol}} = 1.45 \, \text{moles} \][/tex]
Step 2: Determine the limiting reactant
According to the reaction, 2 moles of [tex]\( \text{H}_2 \)[/tex] react with 1 mole of [tex]\( \text{O}_2 \)[/tex]. Therefore:
- The moles of [tex]\( \text{H}_2 \)[/tex] required for 1.45 moles of [tex]\( \text{O}_2 \)[/tex] is:
[tex]\[ 1.45 \, \text{moles of } \text{O}_2 \times 2 = 2.90 \, \text{moles of } \text{H}_2 \][/tex]
Since we only have 1.83 moles of [tex]\( \text{H}_2 \)[/tex], [tex]\( \text{H}_2 \)[/tex] is the limiting reactant.
Step 3: Calculate the moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]) produced
Since [tex]\( \text{H}_2 \)[/tex] is the limiting reactant:
- According to the equation, 2 moles of [tex]\( \text{H}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Thus, 1.83 moles of [tex]\( \text{H}_2 \)[/tex] will produce 1.83 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
Step 4: Calculate the grams of water ([tex]\( \text{H}_2\text{O} \)[/tex]) produced
Given that the molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is 18.02 g/mol:
- Mass of [tex]\( \text{H}_2\text{O} = 1.83 \, \text{moles} \times 18.02 \, \text{g/mol} = 33.01 \, \text{grams} \)[/tex]
Thus, the amount of water produced in this reaction is 33.01 grams.
[tex]\[ 2 \, \text{H}_2 + \text{O}_2 \rightarrow 2 \, \text{H}_2\text{O} \][/tex]
Step 1: Calculate the moles of hydrogen ([tex]\( \text{H}_2 \)[/tex]) and oxygen ([tex]\( \text{O}_2 \)[/tex])
- Hydrogen ([tex]\( \text{H}_2 \)[/tex]):
Given mass of [tex]\( \text{H}_2 = 3.7 \)[/tex] grams
Molar mass of [tex]\( \text{H}_2 = 2.02 \)[/tex] g/mol
[tex]\[ \text{Moles of } \text{H}_2 = \frac{3.7 \, \text{g}}{2.02 \, \text{g/mol}} = 1.83 \, \text{moles} \][/tex]
- Oxygen ([tex]\( \text{O}_2 \)[/tex]):
Given mass of [tex]\( \text{O}_2 = 46.5 \)[/tex] grams
Molar mass of [tex]\( \text{O}_2 = 32 \)[/tex] g/mol
[tex]\[ \text{Moles of } \text{O}_2 = \frac{46.5 \, \text{g}}{32 \, \text{g/mol}} = 1.45 \, \text{moles} \][/tex]
Step 2: Determine the limiting reactant
According to the reaction, 2 moles of [tex]\( \text{H}_2 \)[/tex] react with 1 mole of [tex]\( \text{O}_2 \)[/tex]. Therefore:
- The moles of [tex]\( \text{H}_2 \)[/tex] required for 1.45 moles of [tex]\( \text{O}_2 \)[/tex] is:
[tex]\[ 1.45 \, \text{moles of } \text{O}_2 \times 2 = 2.90 \, \text{moles of } \text{H}_2 \][/tex]
Since we only have 1.83 moles of [tex]\( \text{H}_2 \)[/tex], [tex]\( \text{H}_2 \)[/tex] is the limiting reactant.
Step 3: Calculate the moles of water ([tex]\( \text{H}_2\text{O} \)[/tex]) produced
Since [tex]\( \text{H}_2 \)[/tex] is the limiting reactant:
- According to the equation, 2 moles of [tex]\( \text{H}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Thus, 1.83 moles of [tex]\( \text{H}_2 \)[/tex] will produce 1.83 moles of [tex]\( \text{H}_2\text{O} \)[/tex].
Step 4: Calculate the grams of water ([tex]\( \text{H}_2\text{O} \)[/tex]) produced
Given that the molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is 18.02 g/mol:
- Mass of [tex]\( \text{H}_2\text{O} = 1.83 \, \text{moles} \times 18.02 \, \text{g/mol} = 33.01 \, \text{grams} \)[/tex]
Thus, the amount of water produced in this reaction is 33.01 grams.