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To determine the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced when 10.0 grams of [tex]\( \text{H}_2 \)[/tex] reacts completely with 80.0 grams of [tex]\( \text{O}_2 \)[/tex], follow these steps:
1. Calculate the moles of [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{O}_2 \)[/tex] given their masses and molar masses.
- Molar mass of [tex]\( \text{H}_2 \)[/tex] (Hydrogen gas) is [tex]\( 2.016 \, \text{g/mol} \)[/tex].
- Molar mass of [tex]\( \text{O}_2 \)[/tex] (Oxygen gas) is [tex]\( 32.00 \, \text{g/mol} \)[/tex].
For [tex]\( 10.0 \, \text{g} \)[/tex] of [tex]\( \text{H}_2 \)[/tex]:
[tex]\[ \text{Moles of } H_2 = \frac{10.0 \, \text{g}}{2.016 \, \text{g/mol}} = 4.9603174603174605 \, \text{mol} \][/tex]
For [tex]\( 80.0 \, \text{g} \)[/tex] of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{80.0 \, \text{g}}{32.00 \, \text{g/mol}} = 2.5 \, \text{mol} \][/tex]
2. Use the balanced chemical equation to determine the limiting reactant.
The balanced reaction equation is:
[tex]\[ 2 H_2 + O_2 \rightarrow 2 H_2O \][/tex]
According to the equation, 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 to react with [tex]\( \text{O}_2 \)[/tex] are:
[tex]\[ \text{Moles of } H_2 \text{ required} = 2 \times (\text{Moles of } O_2) = 2 \times 2.5 = 5.0 \, \text{mol} \][/tex]
However, we only have [tex]\( 4.9603174603174605 \)[/tex] moles of [tex]\( \text{H}_2 \)[/tex], which means [tex]\( \text{H}_2 \)[/tex] is the limiting reactant.
3. Calculate the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced.
From the balanced equation:
[tex]\[ 2 H_2 \rightarrow 2 H_2O \][/tex]
This implies the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced are equal to the moles of [tex]\( \text{H}_2 \)[/tex] used. Therefore, the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced are:
[tex]\[ \text{Moles of } H_2O = \frac{\text{Moles of } H_2}{2} \times 2 = 4.9603174603174605 \, \text{mol} \][/tex]
4. Calculate the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced.
- The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] (Water) is [tex]\( 18.016 \, \text{g/mol} \)[/tex].
Therefore, the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced is:
[tex]\[ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar mass of } H_2O = 4.9603174603174605 \, \text{mol} \times 18.016 \, \text{g/mol} = 89.36507936507935 \, \text{g} \][/tex]
Given the choices, the closest and most reasonable matching option is [tex]\( 90.0 \, \text{g} \)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{90.0 \, \text{g}} \][/tex]
1. Calculate the moles of [tex]\( \text{H}_2 \)[/tex] and [tex]\( \text{O}_2 \)[/tex] given their masses and molar masses.
- Molar mass of [tex]\( \text{H}_2 \)[/tex] (Hydrogen gas) is [tex]\( 2.016 \, \text{g/mol} \)[/tex].
- Molar mass of [tex]\( \text{O}_2 \)[/tex] (Oxygen gas) is [tex]\( 32.00 \, \text{g/mol} \)[/tex].
For [tex]\( 10.0 \, \text{g} \)[/tex] of [tex]\( \text{H}_2 \)[/tex]:
[tex]\[ \text{Moles of } H_2 = \frac{10.0 \, \text{g}}{2.016 \, \text{g/mol}} = 4.9603174603174605 \, \text{mol} \][/tex]
For [tex]\( 80.0 \, \text{g} \)[/tex] of [tex]\( \text{O}_2 \)[/tex]:
[tex]\[ \text{Moles of } O_2 = \frac{80.0 \, \text{g}}{32.00 \, \text{g/mol}} = 2.5 \, \text{mol} \][/tex]
2. Use the balanced chemical equation to determine the limiting reactant.
The balanced reaction equation is:
[tex]\[ 2 H_2 + O_2 \rightarrow 2 H_2O \][/tex]
According to the equation, 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 to react with [tex]\( \text{O}_2 \)[/tex] are:
[tex]\[ \text{Moles of } H_2 \text{ required} = 2 \times (\text{Moles of } O_2) = 2 \times 2.5 = 5.0 \, \text{mol} \][/tex]
However, we only have [tex]\( 4.9603174603174605 \)[/tex] moles of [tex]\( \text{H}_2 \)[/tex], which means [tex]\( \text{H}_2 \)[/tex] is the limiting reactant.
3. Calculate the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced.
From the balanced equation:
[tex]\[ 2 H_2 \rightarrow 2 H_2O \][/tex]
This implies the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced are equal to the moles of [tex]\( \text{H}_2 \)[/tex] used. Therefore, the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced are:
[tex]\[ \text{Moles of } H_2O = \frac{\text{Moles of } H_2}{2} \times 2 = 4.9603174603174605 \, \text{mol} \][/tex]
4. Calculate the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced.
- The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] (Water) is [tex]\( 18.016 \, \text{g/mol} \)[/tex].
Therefore, the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced is:
[tex]\[ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar mass of } H_2O = 4.9603174603174605 \, \text{mol} \times 18.016 \, \text{g/mol} = 89.36507936507935 \, \text{g} \][/tex]
Given the choices, the closest and most reasonable matching option is [tex]\( 90.0 \, \text{g} \)[/tex].
Hence, the correct answer is:
[tex]\[ \boxed{90.0 \, \text{g}} \][/tex]
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