A student reacts 832 mL of 5.43 M NazCO3 with excess AICI, in the lab, according to the balanced chemical equation below, and finds that she can produce 0.72 mole of Al2 (CO3)3. What is the percent yield for her experiment? 3NA2CO3(aq) + 2AIC13 (aq) → Al2 (CO3)3(s) + 6NaCl(aq) Please express your answer to the second decimal place. Type your answer.

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**Percent Yield Calculation in a Chemical Reaction**

*Problem Description:*

A student reacts 832 mL of 5.43 M Na₂CO₃ with excess AlCl₃ in the lab, according to the balanced chemical equation below. The student measures that she can produce 0.72 moles of Al₂(CO₃)₃. What is the percent yield for her experiment?

**Balanced Chemical Equation:**

\[ 3Na_2CO_3 (aq) + 2AlCl_3 (aq) \rightarrow Al_2(CO_3)_3 (s) + 6NaCl (aq) \]

**Instructions:**

Please express your answer to the second decimal place. 

[Input Box for Answer]

---

### Step-by-Step Solution:

1. **Determine the moles of Na₂CO₃ used:**
   - Volume of Na₂CO₃: 832 mL \( \rightarrow \) 0.832 L
   - Molarity (M) of Na₂CO₃: 5.43 M

\[ \text{Moles of Na₂CO₃} = \text{Volume (L)} \times \text{Molarity} \]
\[ \text{Moles of Na₂CO₃} = 0.832 \, \text{L} \times 5.43 \, \text{M} = 4.51656 \, \text{moles} \]

2. **Using the stoichiometric ratio from the balanced equation:**

   According to the balanced equation:
   \[ 3 \, \text{moles of } Na₂CO₃ \rightarrow 1 \, \text{mole of } Al₂(CO₃)₃ \]

\[ \text{Moles of Al₂(CO₃)₃ (theoretical yield)} = \frac{4.51656 \, \text{moles of Na₂CO₃}}{3} = 1.50552 \, \text{moles} \]

3. **Compare with the actual yield to find the percent yield:**

   Actual yield of Al₂(CO₃)₃: 0.72 moles

\[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}}
Transcribed Image Text:**Percent Yield Calculation in a Chemical Reaction** *Problem Description:* A student reacts 832 mL of 5.43 M Na₂CO₃ with excess AlCl₃ in the lab, according to the balanced chemical equation below. The student measures that she can produce 0.72 moles of Al₂(CO₃)₃. What is the percent yield for her experiment? **Balanced Chemical Equation:** \[ 3Na_2CO_3 (aq) + 2AlCl_3 (aq) \rightarrow Al_2(CO_3)_3 (s) + 6NaCl (aq) \] **Instructions:** Please express your answer to the second decimal place. [Input Box for Answer] --- ### Step-by-Step Solution: 1. **Determine the moles of Na₂CO₃ used:** - Volume of Na₂CO₃: 832 mL \( \rightarrow \) 0.832 L - Molarity (M) of Na₂CO₃: 5.43 M \[ \text{Moles of Na₂CO₃} = \text{Volume (L)} \times \text{Molarity} \] \[ \text{Moles of Na₂CO₃} = 0.832 \, \text{L} \times 5.43 \, \text{M} = 4.51656 \, \text{moles} \] 2. **Using the stoichiometric ratio from the balanced equation:** According to the balanced equation: \[ 3 \, \text{moles of } Na₂CO₃ \rightarrow 1 \, \text{mole of } Al₂(CO₃)₃ \] \[ \text{Moles of Al₂(CO₃)₃ (theoretical yield)} = \frac{4.51656 \, \text{moles of Na₂CO₃}}{3} = 1.50552 \, \text{moles} \] 3. **Compare with the actual yield to find the percent yield:** Actual yield of Al₂(CO₃)₃: 0.72 moles \[ \text{Percent Yield} = \left( \frac{\text{Actual Yield}}{\text{Theoretical Yield}}
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