Answer :
To determine how many grams of water ([tex]\(\text{H}_2\text{O}\)[/tex]) are produced by reacting 15.8 grams of hydrogen ([tex]\(\text{H}_2\)[/tex]) with excess oxygen ([tex]\(\text{O}_2\)[/tex]), we need to follow several steps.
### Step 1: Calculate the moles of [tex]\( \text{H}_2 \)[/tex]
First, we need the molar mass of [tex]\( \text{H}_2 \)[/tex], which is the mass of one mole of [tex]\( \text{H}_2 \)[/tex] molecules. Since [tex]\( \text{H}_2 \)[/tex] consists of two hydrogen atoms and the atomic mass of hydrogen is approximately 1.008 g/mol, the molar mass of [tex]\( \text{H}_2 \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{H}_2 = 2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol} \][/tex]
Using the molar mass, we can determine the number of moles of [tex]\( \text{H}_2 \)[/tex] present in 15.8 grams:
[tex]\[ \text{Moles of } \text{H}_2 = \frac{\text{Mass of } \text{H}_2}{\text{Molar mass of } \text{H}_2} = \frac{15.8 \text{ g}}{2.016 \text{ g/mol}} \approx 7.837 \text{ moles} \][/tex]
### Step 2: Use the stoichiometry of the reaction
The balanced chemical equation for the reaction is:
[tex]\[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \][/tex]
According to the stoichiometry of the balanced equation, 2 moles of [tex]\( \text{H}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced is equal to the moles of [tex]\( \text{H}_2 \)[/tex] used:
[tex]\[ \text{Moles of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2 = 7.837 \text{ moles} \][/tex]
### Step 3: Calculate the mass of [tex]\( \text{H}_2\text{O} \)[/tex]
Finally, we need the molar mass of [tex]\( \text{H}_2\text{O} \)[/tex]. Water consists of two hydrogen atoms and one oxygen atom. The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{H}_2\text{O} = 2 \times 1.008 \text{ g/mol} + 16.00 \text{ g/mol} = 18.016 \text{ g/mol} \][/tex]
Using this molar mass, we can find the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced:
[tex]\[ \text{Mass of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2\text{O} \times \text{Molar mass of } \text{H}_2\text{O} = 7.837 \text{ moles} \times 18.016 \text{ g/mol} \approx 141.189 \text{ g} \][/tex]
Therefore, the mass of water produced by reacting 15.8 grams of hydrogen ([tex]\(\text{H}_2\)[/tex]) with excess oxygen is approximately 141 grams.
### Answer:
D. [tex]\( 141 \)[/tex] g
### Step 1: Calculate the moles of [tex]\( \text{H}_2 \)[/tex]
First, we need the molar mass of [tex]\( \text{H}_2 \)[/tex], which is the mass of one mole of [tex]\( \text{H}_2 \)[/tex] molecules. Since [tex]\( \text{H}_2 \)[/tex] consists of two hydrogen atoms and the atomic mass of hydrogen is approximately 1.008 g/mol, the molar mass of [tex]\( \text{H}_2 \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{H}_2 = 2 \times 1.008 \text{ g/mol} = 2.016 \text{ g/mol} \][/tex]
Using the molar mass, we can determine the number of moles of [tex]\( \text{H}_2 \)[/tex] present in 15.8 grams:
[tex]\[ \text{Moles of } \text{H}_2 = \frac{\text{Mass of } \text{H}_2}{\text{Molar mass of } \text{H}_2} = \frac{15.8 \text{ g}}{2.016 \text{ g/mol}} \approx 7.837 \text{ moles} \][/tex]
### Step 2: Use the stoichiometry of the reaction
The balanced chemical equation for the reaction is:
[tex]\[ 2 \text{H}_2 + \text{O}_2 \rightarrow 2 \text{H}_2\text{O} \][/tex]
According to the stoichiometry of the balanced equation, 2 moles of [tex]\( \text{H}_2 \)[/tex] produce 2 moles of [tex]\( \text{H}_2\text{O} \)[/tex]. Therefore, the moles of [tex]\( \text{H}_2\text{O} \)[/tex] produced is equal to the moles of [tex]\( \text{H}_2 \)[/tex] used:
[tex]\[ \text{Moles of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2 = 7.837 \text{ moles} \][/tex]
### Step 3: Calculate the mass of [tex]\( \text{H}_2\text{O} \)[/tex]
Finally, we need the molar mass of [tex]\( \text{H}_2\text{O} \)[/tex]. Water consists of two hydrogen atoms and one oxygen atom. The molar mass of [tex]\( \text{H}_2\text{O} \)[/tex] is:
[tex]\[ \text{Molar mass of } \text{H}_2\text{O} = 2 \times 1.008 \text{ g/mol} + 16.00 \text{ g/mol} = 18.016 \text{ g/mol} \][/tex]
Using this molar mass, we can find the mass of [tex]\( \text{H}_2\text{O} \)[/tex] produced:
[tex]\[ \text{Mass of } \text{H}_2\text{O} = \text{Moles of } \text{H}_2\text{O} \times \text{Molar mass of } \text{H}_2\text{O} = 7.837 \text{ moles} \times 18.016 \text{ g/mol} \approx 141.189 \text{ g} \][/tex]
Therefore, the mass of water produced by reacting 15.8 grams of hydrogen ([tex]\(\text{H}_2\)[/tex]) with excess oxygen is approximately 141 grams.
### Answer:
D. [tex]\( 141 \)[/tex] g
