Aluminium has a density of 2.70 g/cm. What quantity in moles of aluminum are in a 13.2 cm block of the metal substance? JAT mol Al 0.100 mol Al 2.70 cm3 26.98 JAT 2.70 g Al 1 mol Al = 0.100 mol AI cm3 26.98 g AI ADD FACTOR DELETE ANSWER RESET グ 13.2 35.6 0.489 1 1.32 26.98 6.022 x 10 962 0.100 2.70 g/cm g Al molecules Al mol Al g Al/mol cm³
Aluminium has a density of 2.70 g/cm. What quantity in moles of aluminum are in a 13.2 cm block of the metal substance? JAT mol Al 0.100 mol Al 2.70 cm3 26.98 JAT 2.70 g Al 1 mol Al = 0.100 mol AI cm3 26.98 g AI ADD FACTOR DELETE ANSWER RESET グ 13.2 35.6 0.489 1 1.32 26.98 6.022 x 10 962 0.100 2.70 g/cm g Al molecules Al mol Al g Al/mol cm³
Chapter5: Chemical Reactions
Section: Chapter Questions
Problem 5.40E
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![### How to Calculate Moles of Aluminum in a Given Volume
Aluminum has a density of 2.70 g/cm³. We want to determine the quantity, in moles, of aluminum present in a 13.2 cm³ block of the metal.
#### Calculation Steps
1. **Given Data:**
- Density of aluminum: 2.70 g/cm³
- Volume of aluminum block: 13.2 cm³
2. **Formula Used:**
\[
\text{Moles of Al} = \left( \frac{\text{Volume of Al block} \times \text{Density of Al}}{\text{Molar mass of Al}} \right)
\]
3. **Substitute the Values:**
\[
\text{Moles of Al} = \left( \frac{13.2 \, \text{cm}^3 \times 2.70 \, \text{g/cm}^3}{26.98 \, \text{g/mol}} \right)
\]
4. **Calculate:**
- The multiplication of volume and density gives mass in grams.
- This equals:
\[
\left( \frac{35.64 \, \text{g}}{26.98 \, \text{g/mol}} \right) = 0.100 \, \text{mol Al}
\]
#### Visual Representation
- The image includes buttons and inputs allowing students to interactively change values.
- Key values like density (2.70), molar mass of aluminum (26.98 g/mol), and volume (13.2 cm³) are highlighted.
- The result box shows the calculated moles: 0.100 mol Al.
#### Tools & Features
- **Add Factor & Delete Tools**: Allow manipulation of calculation steps.
- **Answer Box**: Shows final calculated value in moles.
- **Reset Function**: Clears inputs for new data entry.
- **Interactive Input Keys**: Include numbers and units for ease of recalculation.
This interactive approach aids in visual learning, helping students understand the relationship between density, mass, and moles.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F977c56cd-e376-4ff8-b1da-b449ec52f3a5%2F80cb8edb-0d3a-4e4b-9eae-6b7a987c737e%2Fgxrqax_processed.png&w=3840&q=75)
Transcribed Image Text:### How to Calculate Moles of Aluminum in a Given Volume
Aluminum has a density of 2.70 g/cm³. We want to determine the quantity, in moles, of aluminum present in a 13.2 cm³ block of the metal.
#### Calculation Steps
1. **Given Data:**
- Density of aluminum: 2.70 g/cm³
- Volume of aluminum block: 13.2 cm³
2. **Formula Used:**
\[
\text{Moles of Al} = \left( \frac{\text{Volume of Al block} \times \text{Density of Al}}{\text{Molar mass of Al}} \right)
\]
3. **Substitute the Values:**
\[
\text{Moles of Al} = \left( \frac{13.2 \, \text{cm}^3 \times 2.70 \, \text{g/cm}^3}{26.98 \, \text{g/mol}} \right)
\]
4. **Calculate:**
- The multiplication of volume and density gives mass in grams.
- This equals:
\[
\left( \frac{35.64 \, \text{g}}{26.98 \, \text{g/mol}} \right) = 0.100 \, \text{mol Al}
\]
#### Visual Representation
- The image includes buttons and inputs allowing students to interactively change values.
- Key values like density (2.70), molar mass of aluminum (26.98 g/mol), and volume (13.2 cm³) are highlighted.
- The result box shows the calculated moles: 0.100 mol Al.
#### Tools & Features
- **Add Factor & Delete Tools**: Allow manipulation of calculation steps.
- **Answer Box**: Shows final calculated value in moles.
- **Reset Function**: Clears inputs for new data entry.
- **Interactive Input Keys**: Include numbers and units for ease of recalculation.
This interactive approach aids in visual learning, helping students understand the relationship between density, mass, and moles.
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