To determine how many grams of sodium sulfate (Na2SO4) can be produced from 50.0 grams of sodium hydroxide (NaOH), we first need to use the balanced chemical equation provided:
\[ 2 \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2 \text{H}_2\text{O} \]
Step 1: Calculate the number of moles of NaOH
First, we need the molar mass of sodium hydroxide (NaOH):
- Sodium (Na): 22.99 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.01 g/mol
Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Now, we can calculate the number of moles of NaOH in 50.0 grams:
\[ \text{moles of NaOH} = \frac{\text{mass}}{\text{molar mass}} = \frac{50.0 \text{ g}}{40.00 \text{ g/mol}} = 1.25 \text{ moles of NaOH} \]
Step 2: Use the stoichiometry of the reaction
From the balanced equation, the molar ratio of NaOH to Na2SO4 is 2:1. This means that 2 moles of NaOH produce 1 mole of Na2SO4. Therefore, we can calculate the moles of Na2SO4 produced from the moles of NaOH:
\[ \text{moles of Na}_2\text{SO}_4 = \frac{1.25 \text{ moles NaOH}}{2} = 0.625 \text{ moles of Na}_2\text{SO}_4 \]
Step 3: Calculate the mass of Na2SO4 produced
Next, we need to find the molar mass of sodium sulfate (Na2SO4):
- Sodium (Na): 22.99 g/mol
- Sulfur (S): 32.07 g/mol
- Oxygen (O): 16.00 g/mol (4 oxygen atoms)
Molar mass of Na2SO4 = (2 × 22.99) + 32.07 + (4 × 16.00) = 45.98 + 32.07 + 64.00 = 142.05 g/mol
Now we can calculate the mass of Na2SO4 produced:
\[ \text{mass of Na}_2\text{SO}_4 = \text{moles} \times \text{molar mass} = 0.625 \text{ moles} \times 142.05 \text{ g/mol} \approx 88.78 \text{ g} \]
Conclusion
You can produce approximately 88.78 grams of sodium sulfate (Na2SO4) from 50.0 grams of sodium hydroxide (NaOH) with an excess of sulfuric acid.
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