Questions: Determine the number of moles in each of the quantities below. a) 25 g NaCl b) 125 g H2SO4 c) 3,100 g KMnO4 d) 74 g KCl e) 35 g CuSO4

Determine the number of moles in each of the quantities below. a) 25 g NaCl b) 125 g H2SO4 c) 3,100 g KMnO4 d) 74 g KCl e) 35 g CuSO4
Transcript text: 2. Determine the number of moles in each of the quantities below. a) 25 g NaCl b) $125 \mathrm{~g} \mathrm{H}_{2} \mathrm{SO}_{4}$ c) $3,100 . \mathrm{g} \mathrm{KMnO}_{4}$ d) 74 g KCl e) $35 \mathrm{~g} \mathrm{CuSO}_{4}$
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Solution

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Solution Steps

Step 1: Calculate Molar Masses

First, we need to calculate the molar masses of each compound.

  • For NaCl:

    • Na: \(22.99 \, \text{g/mol}\)
    • Cl: \(35.45 \, \text{g/mol}\)
    • Molar mass of NaCl: \(22.99 + 35.45 = 58.44 \, \text{g/mol}\)

  • For \( \mathrm{H}_{2} \mathrm{SO}_{4} \):

    • H: \(1.008 \, \text{g/mol} \times 2 = 2.016 \, \text{g/mol}\)
    • S: \(32.06 \, \text{g/mol}\)
    • O: \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)
    • Molar mass of \( \mathrm{H}_{2} \mathrm{SO}_{4} \): \(2.016 + 32.06 + 64.00 = 98.08 \, \text{g/mol}\)

  • For \( \mathrm{KMnO}_{4} \):

    • K: \(39.10 \, \text{g/mol}\)
    • Mn: \(54.94 \, \text{g/mol}\)
    • O: \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)
    • Molar mass of \( \mathrm{KMnO}_{4} \): \(39.10 + 54.94 + 64.00 = 158.04 \, \text{g/mol}\)

  • For KCl:

    • K: \(39.10 \, \text{g/mol}\)
    • Cl: \(35.45 \, \text{g/mol}\)
    • Molar mass of KCl: \(39.10 + 35.45 = 74.55 \, \text{g/mol}\)

  • For \( \mathrm{CuSO}_{4} \):

    • Cu: \(63.55 \, \text{g/mol}\)
    • S: \(32.06 \, \text{g/mol}\)
    • O: \(16.00 \, \text{g/mol} \times 4 = 64.00 \, \text{g/mol}\)
    • Molar mass of \( \mathrm{CuSO}_{4} \): \(63.55 + 32.06 + 64.00 = 159.61 \, \text{g/mol}\)
Step 2: Calculate Number of Moles

Next, we use the formula: \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

  • For 25 g NaCl: \[ \text{Number of moles} = \frac{25 \, \text{g}}{58.44 \, \text{g/mol}} = 0.4279 \, \text{moles} \]

  • For 125 g \( \mathrm{H}_{2} \mathrm{SO}_{4} \): \[ \text{Number of moles} = \frac{125 \, \text{g}}{98.08 \, \text{g/mol}} = 1.274 \, \text{moles} \]

  • For 3100 g \( \mathrm{KMnO}_{4} \): \[ \text{Number of moles} = \frac{3100 \, \text{g}}{158.04 \, \text{g/mol}} = 19.62 \, \text{moles} \]

Final Answer

\[ \boxed{\text{a) } 0.4279 \, \text{moles}} \] \[ \boxed{\text{b) } 1.274 \, \text{moles}} \] \[ \boxed{\text{c) } 19.62 \, \text{moles}} \]

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