2.41 g of Carbon dioxide gas and 3.50 grams of Sodium hydroxide are re lab, 4.75 grams of Sodium carbonate is produced. Co2(g) + 2 NaOH(ag) → Na2CO3(ag) + H20(1) i. Find the limiting reactant ii. Find the theoretical yield of Na2CO3

Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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Balance the following chemical reaction, and complete the stoichiometry problem:

 

**Chemical Reaction Problem:**

**Description:**
Carbon dioxide reacts with sodium hydroxide to form sodium carbonate and water. When 2.41 g of Carbon dioxide gas and 3.50 grams of Sodium hydroxide are reacted together in a lab, 4.75 grams of Sodium carbonate is produced.

**Reaction Equation:**
\[
\text{CO}_2(\text{g}) + 2 \, \text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l})
\]

---

**Questions to Solve:**

**i.** Find the limiting reactant.

**ii.** Find the theoretical yield of Na$_2$CO$_3$.

**iii.** Find the % yield of Na$_2$CO$_3$ using the amount given in the problem.

---

Please solve each part step by step for a thorough understanding of chemical stoichiometry.

**i. Finding the Limiting Reactant:**

1. **Molar Mass Calculation**:
   - Molar mass of CO$_2$: \( 12.01 \, (\text{C}) + 2 \times 16.00 \, (\text{O}) = 44.01 \, \text{g/mol} \)
   - Molar mass of NaOH: \( 22.99 \, (\text{Na}) + 15.99 \, (\text{O}) + 1.01 \, (\text{H}) = 39.99 \, \text{g/mol} \)

2. **Moles of Reactants**:
   - Moles of CO$_2$: \( \frac{2.41 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0547 \, \text{moles} \)
   - Moles of NaOH: \( \frac{3.50 \, \text{g}}{39.99 \, \text{g/mol}} = 0.0875 \, \text{moles} \)

3. **Reaction Stoichiometry**:
   - From the balanced equation: 1 mole of CO$_2$ reacts with 2 moles of NaOH.
   - Required NaOH for 0.054
Transcribed Image Text:**Chemical Reaction Problem:** **Description:** Carbon dioxide reacts with sodium hydroxide to form sodium carbonate and water. When 2.41 g of Carbon dioxide gas and 3.50 grams of Sodium hydroxide are reacted together in a lab, 4.75 grams of Sodium carbonate is produced. **Reaction Equation:** \[ \text{CO}_2(\text{g}) + 2 \, \text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{CO}_3(\text{aq}) + \text{H}_2\text{O}(\text{l}) \] --- **Questions to Solve:** **i.** Find the limiting reactant. **ii.** Find the theoretical yield of Na$_2$CO$_3$. **iii.** Find the % yield of Na$_2$CO$_3$ using the amount given in the problem. --- Please solve each part step by step for a thorough understanding of chemical stoichiometry. **i. Finding the Limiting Reactant:** 1. **Molar Mass Calculation**: - Molar mass of CO$_2$: \( 12.01 \, (\text{C}) + 2 \times 16.00 \, (\text{O}) = 44.01 \, \text{g/mol} \) - Molar mass of NaOH: \( 22.99 \, (\text{Na}) + 15.99 \, (\text{O}) + 1.01 \, (\text{H}) = 39.99 \, \text{g/mol} \) 2. **Moles of Reactants**: - Moles of CO$_2$: \( \frac{2.41 \, \text{g}}{44.01 \, \text{g/mol}} = 0.0547 \, \text{moles} \) - Moles of NaOH: \( \frac{3.50 \, \text{g}}{39.99 \, \text{g/mol}} = 0.0875 \, \text{moles} \) 3. **Reaction Stoichiometry**: - From the balanced equation: 1 mole of CO$_2$ reacts with 2 moles of NaOH. - Required NaOH for 0.054
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