To find the mass of \[ H_2O \] produced, we can use stoichiometry to first determine which reactant is limiting and then calculate the mass of \[ H_2O \] produced.
First, calculate the number of moles of each reactant using their molar masses:
\[
\text{Molar mass of } H_2 = 2 \, \text{g/mol} \\
\text{Moles of } H_2 = \frac{10.0 \, \text{g}}{2 \, \text{g/mol}} = 5.0 \, \text{mol}
\]
\[
\text{Molar mass of } O_2 = 32 \, \text{g/mol} \\
\text{Moles of } O_2 = \frac{80.0 \, \text{g}}{32 \, \text{g/mol}} = 2.5 \, \text{mol}
\]
The balanced equation shows that the mole ratio of \[ H_2 \] to \[ O_2 \] is 2:1. Since we have a 2:1 ratio in the reaction, \[ O_2 \] is the limiting reactant because we only have 2.5 moles of it compared to 5.0 moles of \[ H_2 \].
Next, we calculate the theoretical yield of \[ H_2O \] produced from the limiting reactant:
\[ \text{Molar mass of } H_2O = 18 \, \text{g/mol} \]
\[ \text{Moles of } H_2O = 2 \times 2.5 \, \text{mol} = 5.0 \, \text{mol} \]
\[ \text{Mass of } H_2O = 5.0 \, \text{mol} \times 18 \, \text{g/mol} = 90.0 \, \text{g} \]
Therefore, the mass of \[ H_2O \] produced when 10.0 grams of \[ H_2 \] reacts completely with 80.0 grams of \[ O_2 \] is \[ 90.0 \, g \]. The correct answer is 2. \[ 90.0 \, g \].
Given the balanced equation representing a reaction:
\[ 2 H_2 + O_2 \rightarrow 2 H_2O \]
What is the mass of \[ H_2O \] produced when 10.0 grams of \[ H_2 \] reacts completely with 80.0 grams of \[ O_2 \]?
1. \[ 70.0 \, g \]
2. \[ 90.0 \, g \]
3. \[ 180.0 \, g \]
4. \[ 800.0 \, g \]
1 answer