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.
States of Matter
The substance that constitutes everything in the universe is known as matter. Matter comprises atoms which in turn are composed of electrons, protons, and neutrons. Different atoms combine together to give rise to molecules that act as a foundation for all kinds of substances. There are five states of matter based on their energies of attraction, namely solid, liquid, gases, plasma, and BEC (Bose-Einstein condensates).
Chemical Reactions and Equations
When a chemical species is transformed into another chemical species it is said to have undergone a chemical reaction. It consists of breaking existing bonds and forming new bonds by changing the position of electrons. These reactions are best explained using a chemical equation.
![**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]
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### 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}}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F19b343d3-def3-436d-96c5-7f330515e000%2F210a947e-47b9-4b13-8ca8-6b66d5045c10%2Fagho9u_processed.jpeg&w=3840&q=75)

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