1. What is the maximum amount of sodium carbonate that can be produced from 12.5 grams of $$ \mathrm{CS}_{2} $$ and 14 grams of NaOH ? Identify the limiting and excess reactants for this reaction, $$ 3 \mathrm{CS}_{2}+6 \mathrm{NaOH} \rightarrow 2 \mathrm{Na}_{2} \mathrm{CS}_{3}+1 \mathrm{Na}_{2} \mathrm{CO}_{3}+3 \mathrm{H}_{2} \mathrm{O} $$ I

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To determine the maximum amount of sodium carbonate ($$\mathrm{Na}_2\mathrm{CO}_3$$) that can be produced from 12.5 grams of carbon disulfide ($$\mathrm{CS}_2$$) and 14 grams of sodium hydroxide ($$\mathrm{NaOH}$$), we'll follow these steps:

1. **Calculate the Molar Masses:**
   - **$$\mathrm{CS}_2$$:**
     $$
     \text{C} = 12.01\, \text{g/mol},\ \text{S} = 32.07\, \text{g/mol} \
     M_{\mathrm{CS}_2} = 12.01 + (2 \times 32.07) = 76.15\, \text{g/mol}
     $$
   - **$$\mathrm{NaOH}$$:**
     $$
     \text{Na} = 22.99\, \text{g/mol},\ \text{O} = 16.00\, \text{g/mol},\ \text{H} = 1.01\, \text{g/mol} \
     M_{\mathrm{NaOH}} = 22.99 + 16.00 + 1.01 = 40.00\, \text{g/mol}
     $$
   - **$$\mathrm{Na}_2\mathrm{CO}_3$$:**
     $$
     M_{\mathrm{Na}_2\mathrm{CO}_3} = (2 \times 22.99) + 12.01 + (3 \times 16.00) = 105.99\, \text{g/mol}
     $$

2. **Determine the Number of Moles:**
   - **$$\mathrm{CS}_2$$:**
     $$
     \text{Moles of } \mathrm{CS}_2 = \frac{12.5\, \text{g}}{76.15\, \text{g/mol}} \approx 0.164\, \text{mol}
     $$
   - **$$\mathrm{NaOH}$$:**
     $$
     \text{Moles of } \mathrm{NaOH} = \frac{14\, \text{g}}{40.00\, \text{g/mol}} = 0.35\, \text{mol}
     $$

3. **Identify the Limiting Reactant:**
   The balanced chemical equation is:
   $$
   3\, \mathrm{CS}_2 + 6\, \mathrm{NaOH} \rightarrow 2\, \mathrm{Na}_2\mathrm{CS}_3 + \mathrm{Na}_2\mathrm{CO}_3 + 3\, \mathrm{H}_2\mathrm{O}
   $$
   - **Mole Ratio Needed:**
     $$
     3\, \mathrm{CS}_2 : 6\, \mathrm{NaOH} \Rightarrow 1\, \mathrm{CS}_2 : 2\, \mathrm{NaOH}
     $$
   - **Required $$\mathrm{NaOH}$$ for Available $$\mathrm{CS}_2$$:**
     $$
     0.164\, \text{mol}\, \mathrm{CS}_2 \times \frac{2\, \mathrm{NaOH}}{1\, \mathrm{CS}_2} = 0.328\, \text{mol}\, \mathrm{NaOH}
     $$
   - **Comparison:**
     $$
     0.35\, \text{mol}\, \mathrm{NaOH} > 0.328\, \text{mol}\, \mathrm{NaOH} \quad \Rightarrow \quad \mathrm{CS}_2\ \text{is the limiting reactant}
     $$
     $$
     \text{Excess Reactant: }\mathrm{NaOH}
     $$

4. **Calculate the Maximum Amount of $$\mathrm{Na}_2\mathrm{CO}_3$$ Produced:**
   - **From the Balanced Equation:**
     $$
     3\, \mathrm{CS}_2 \rightarrow 1\, \mathrm{Na}_2\mathrm{CO}_3 \Rightarrow 1\, \mathrm{CS}_2 \rightarrow \frac{1}{3}\, \mathrm{Na}_2\mathrm{CO}_3
     $$
   - **Moles of $$\mathrm{Na}_2\mathrm{CO}_3$$:**
     $$
     0.164\, \text{mol}\, \mathrm{CS}_2 \times \frac{1}{3} = 0.0547\, \text{mol}\, \mathrm{Na}_2\mathrm{CO}_3
     $$
   - **Mass of $$\mathrm{Na}_2\mathrm{CO}_3$$:**
     $$
     0.0547\, \text{mol} \times 105.99\, \text{g/mol} \approx 5.8\, \text{g}
     $$

**Conclusion:**
- **Maximum $$\mathrm{Na}_2\mathrm{CO}_3$$ Produced:** Approximately **5.8 grams**
- **Limiting Reactant:** **$$\mathrm{CS}_2$$**
- **Excess Reactant:** **$$\mathrm{NaOH}$$**
The maximum amount of sodium carbonate that can be produced is approximately 5.8 grams. Carbon disulfide ($$\mathrm{CS}_2$$) is the limiting reactant, and sodium hydroxide ($$\mathrm{NaOH}$$) is the excess reactant.

What is the maximum amount of sodium carbonate that can be produced from 12.5 grams of
and 14 grams of NaOH ? Identify the limiting and excess reactants for this reaction,

I

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What is the maximum amount of sodium carbonate that can be produced from 12.5 grams of and 14 grams of NaOH ? Identify the limiting and excess reactants for this reaction, I