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examples of scaled mole conversions?

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Scaled mole conversions involve concepts of converting between different quantities in chemistry, especially when dealing with moles. Here are a few examples that illustrate this idea:

Example 1: Molar Mass Conversion

Problem: How many grams are in 3 moles of water (H₂O)?

Solution:

  1. Calculate the molar mass of water (H₂O):

    • Hydrogen (H) = 1 g/mol × 2 = 2 g/mol
    • Oxygen (O) = 16 g/mol × 1 = 16 g/mol
    • Total molar mass of H₂O = 2 + 16 = 18 g/mol

  2. Use the molar mass to convert moles to grams: \[ \text{Mass} = \text{moles} \times \text{molar mass} = 3 , \text{moles} \times 18 , \text{g/g/mol} = 54 , \text{g} \]

Example 2: Volume of Gas at STP

Problem: How many liters of gas are produced if 2 moles of carbon dioxide (CO₂) are generated at standard temperature and pressure (STP)?

Solution: At STP, 1 mole of an ideal gas occupies 22.4 liters.

  1. Use the molar volume to convert moles to liters: \[ \text{Volume} = \text{moles} \times 22.4 , \text{L/mol} = 2 , \text{moles} \times 22.4 , \text{L/mol} = 44.8 , \text{L} \]

Example 3: Stoichiometric Calculations in a Reaction

Problem: Given the reaction \( 2 , \text{H}_2 + \text{O}_2 \rightarrow 2 , \text{H}_2\text{O} \), how many moles of \( \text{H}_2\text{O} \) can be produced from 4 moles of \( \text{H}_2 \)?

Solution: From the balanced equation, 2 moles of \( \text{H}_2 \) produce 2 moles of \( \text{H}_2\text{O} \).

  1. Use the mole ratio from the reaction: \[ \frac{2 , \text{mol } \text{H}_2\text{O}}{2 , \text{mol } \text{H}_2} = 1 , \text{mol } \text{H}_2\text{O}/ \text{mol } \text{H}_2 \]

  2. Calculate the number of moles of \( \text{H}_2\text{O} \): \[ \text{Moles of } \text{H}_2\text{O} = 4 , \text{moles } \text{H}_2 \times \frac{2 , \text{moles } \text{H}_2\text{O}}{2 , \text{moles } \text{H}_2} = 4 , \text{moles of } \text{H}_2\text{O} \]

Example 4: Concentration Molarity

Problem: If you have a 2 M (molar) solution of sodium chloride (NaCl), how many moles are in 500 mL of this solution?

Solution:

  1. Convert volume to liters: \[ 500 , \text{mL} = 0.500 , \text{L} \]

  2. Use the molarity formula: \[ \text{Moles} = \text{Molarity} \times \text{Volume} = 2 , \text{mol/L} \times 0.500 , \text{L} = 1 , \text{mol} \]

These examples demonstrate various scenarios in which you can scale mole conversions in chemical calculations.