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To solve the problem of determining the number of moles of sodium hydroxide (NaOH) produced when 2 moles of sodium (Na) and 3 moles of water (H₂O) react, we start with the balanced chemical equation for the reaction:
[tex]\[ 2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2 \][/tex]
### Step-by-Step Solution:
#### Step 1: Identify Stoichiometric Ratios
From the balanced equation, the stoichiometric coefficients for the reactants and products are:
- Sodium (Na) has a coefficient of 2.
- Water (H₂O) has a coefficient of 2.
- Sodium hydroxide (NaOH) has a coefficient of 2.
This tells us that:
- 2 moles of Na react with 2 moles of H₂O to produce 2 moles of NaOH.
#### Step 2: Calculate Moles of NaOH Produced from Each Reactant
Given the stoichiometric ratios and the provided amounts of reactants, we calculate the moles of NaOH that can be produced from each reactant.
From Sodium (Na):
The amount of NaOH that can be produced from sodium is determined by:
[tex]\[ \text{moles of Na} \times \left( \frac{\text{stoichiometric coefficient of NaOH}}{\text{stoichiometric coefficient of Na}} \right) \][/tex]
[tex]\[ = 2 \text{ moles of Na} \times \left( \frac{2 \text{ moles of NaOH}}{2 \text{ moles of Na}} \right) \][/tex]
[tex]\[ = 2.0 \text{ moles of NaOH} \][/tex]
From Water (H₂O):
The amount of NaOH that can be produced from water is determined by:
[tex]\[ \text{moles of H₂O} \times \left( \frac{\text{stoichiometric coefficient of NaOH}}{\text{stoichiometric coefficient of H₂O}} \right) \][/tex]
[tex]\[ = 3 \text{ moles of H₂O} \times \left( \frac{2 \text{ moles of NaOH}}{2 \text{ moles of H₂O}} \right) \][/tex]
[tex]\[ = 3.0 \text{ moles of NaOH} \][/tex]
#### Step 3: Determine the Limiting Reactant
The limiting reactant is the reactant that produces the smaller amount of NaOH.
Comparing the amounts:
- Sodium (Na) produces 2.0 moles of NaOH.
- Water (H₂O) produces 3.0 moles of NaOH.
Thus, sodium (Na) is the limiting reactant because it produces the lesser amount of NaOH.
#### Step 4: Determine the Number of Moles of NaOH Produced
Since the limiting reactant determines the amount of product formed, the actual number of moles of NaOH produced is:
[tex]\[ \text{moles of NaOH produced} = \text{2.0 moles of NaOH} \][/tex]
Therefore, when 2 mol of sodium and 3 mol of water react, the number of moles of sodium hydroxide (NaOH) produced is 2.0 moles.
[tex]\[ 2 \text{Na} + 2 \text{H}_2\text{O} \rightarrow 2 \text{NaOH} + \text{H}_2 \][/tex]
### Step-by-Step Solution:
#### Step 1: Identify Stoichiometric Ratios
From the balanced equation, the stoichiometric coefficients for the reactants and products are:
- Sodium (Na) has a coefficient of 2.
- Water (H₂O) has a coefficient of 2.
- Sodium hydroxide (NaOH) has a coefficient of 2.
This tells us that:
- 2 moles of Na react with 2 moles of H₂O to produce 2 moles of NaOH.
#### Step 2: Calculate Moles of NaOH Produced from Each Reactant
Given the stoichiometric ratios and the provided amounts of reactants, we calculate the moles of NaOH that can be produced from each reactant.
From Sodium (Na):
The amount of NaOH that can be produced from sodium is determined by:
[tex]\[ \text{moles of Na} \times \left( \frac{\text{stoichiometric coefficient of NaOH}}{\text{stoichiometric coefficient of Na}} \right) \][/tex]
[tex]\[ = 2 \text{ moles of Na} \times \left( \frac{2 \text{ moles of NaOH}}{2 \text{ moles of Na}} \right) \][/tex]
[tex]\[ = 2.0 \text{ moles of NaOH} \][/tex]
From Water (H₂O):
The amount of NaOH that can be produced from water is determined by:
[tex]\[ \text{moles of H₂O} \times \left( \frac{\text{stoichiometric coefficient of NaOH}}{\text{stoichiometric coefficient of H₂O}} \right) \][/tex]
[tex]\[ = 3 \text{ moles of H₂O} \times \left( \frac{2 \text{ moles of NaOH}}{2 \text{ moles of H₂O}} \right) \][/tex]
[tex]\[ = 3.0 \text{ moles of NaOH} \][/tex]
#### Step 3: Determine the Limiting Reactant
The limiting reactant is the reactant that produces the smaller amount of NaOH.
Comparing the amounts:
- Sodium (Na) produces 2.0 moles of NaOH.
- Water (H₂O) produces 3.0 moles of NaOH.
Thus, sodium (Na) is the limiting reactant because it produces the lesser amount of NaOH.
#### Step 4: Determine the Number of Moles of NaOH Produced
Since the limiting reactant determines the amount of product formed, the actual number of moles of NaOH produced is:
[tex]\[ \text{moles of NaOH produced} = \text{2.0 moles of NaOH} \][/tex]
Therefore, when 2 mol of sodium and 3 mol of water react, the number of moles of sodium hydroxide (NaOH) produced is 2.0 moles.
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