Answer :
First, recognize the chemical equation given for the decomposition of water:
[tex]\[ 2 H_2O \rightarrow 2 H_2 + O_2 \][/tex]
This balanced reaction shows that 2 moles of water (\(H_2O\)) produce 1 mole of oxygen gas (\(O_2\)).
Given:
- The mass of oxygen gas (\(O_2\)) produced is \(50.00\) grams.
- The molar mass of oxygen gas (\(O_2\)) is \(32.00\) grams per mole.
- The molar mass of water (\(H_2O\)) is \(18.02\) grams per mole.
### Step-by-Step Solution:
1. Calculate the moles of \( O_2 \) produced:
[tex]\[ \text{Moles of } O_2 = \frac{\text{Mass of } O_2}{\text{Molar mass of } O_2} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{50.00 \text{ g}}{32.00 \text{ g/mol}} = 1.5625 \text{ moles} \][/tex]
2. Determine the moles of \( H_2O \) required using the stoichiometry of the reaction:
From the balanced equation, 2 moles of \( H_2O \) produce 1 mole of \( O_2 \).
Hence, to find the moles of \( H_2O \) required to produce \(1.5625\) moles of \( O_2 \):
[tex]\[ \text{Moles of } H_2O = 2 \times \text{Moles of } O_2 \][/tex]
[tex]\[ \text{Moles of } H_2O = 2 \times 1.5625 = 3.125 \text{ moles} \][/tex]
3. Calculate the mass of \( H_2O \) required:
[tex]\[ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar mass of } H_2O \][/tex]
[tex]\[ \text{Mass of } H_2O = 3.125 \text{ moles} \times 18.02 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of } H_2O = 56.3125 \text{ grams} \][/tex]
Thus, the mass of \( H_2O \) that must react to produce \(50.00\) grams of \( O_2 \) is approximately \( 56.31 \) grams.
Therefore, the correct answer is:
[tex]\[ \boxed{56.31 \text{ grams}} \][/tex]
[tex]\[ 2 H_2O \rightarrow 2 H_2 + O_2 \][/tex]
This balanced reaction shows that 2 moles of water (\(H_2O\)) produce 1 mole of oxygen gas (\(O_2\)).
Given:
- The mass of oxygen gas (\(O_2\)) produced is \(50.00\) grams.
- The molar mass of oxygen gas (\(O_2\)) is \(32.00\) grams per mole.
- The molar mass of water (\(H_2O\)) is \(18.02\) grams per mole.
### Step-by-Step Solution:
1. Calculate the moles of \( O_2 \) produced:
[tex]\[ \text{Moles of } O_2 = \frac{\text{Mass of } O_2}{\text{Molar mass of } O_2} \][/tex]
[tex]\[ \text{Moles of } O_2 = \frac{50.00 \text{ g}}{32.00 \text{ g/mol}} = 1.5625 \text{ moles} \][/tex]
2. Determine the moles of \( H_2O \) required using the stoichiometry of the reaction:
From the balanced equation, 2 moles of \( H_2O \) produce 1 mole of \( O_2 \).
Hence, to find the moles of \( H_2O \) required to produce \(1.5625\) moles of \( O_2 \):
[tex]\[ \text{Moles of } H_2O = 2 \times \text{Moles of } O_2 \][/tex]
[tex]\[ \text{Moles of } H_2O = 2 \times 1.5625 = 3.125 \text{ moles} \][/tex]
3. Calculate the mass of \( H_2O \) required:
[tex]\[ \text{Mass of } H_2O = \text{Moles of } H_2O \times \text{Molar mass of } H_2O \][/tex]
[tex]\[ \text{Mass of } H_2O = 3.125 \text{ moles} \times 18.02 \text{ g/mol} \][/tex]
[tex]\[ \text{Mass of } H_2O = 56.3125 \text{ grams} \][/tex]
Thus, the mass of \( H_2O \) that must react to produce \(50.00\) grams of \( O_2 \) is approximately \( 56.31 \) grams.
Therefore, the correct answer is:
[tex]\[ \boxed{56.31 \text{ grams}} \][/tex]
