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How many grams of 0.491 wt% aqueous HF are requires to provide a 50% excess to reat with 25.00 mL of 0.0236 M Th4+ by the reaction Th4+ + 4 F- ® ThF4 (s)

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Edufirst

Answer:

  • 14.4 g

Explanation:

1. Chemical equation:

  • Th⁴⁺   +   4F⁻   →   ThF₄(s)

2. Mole ratio

      [tex]\dfrac{4molF^-}{1molTh^{4+}}[/tex]

3. Number of moles of Th⁴⁺

Molarity, M, is equal to the number of moles, n, per liter, V, of solution

  • M = n / V ⇒ n = M × V
  • M = 0.0236
  • n = 0.0236 × 25.0mL × 1 liter / 1,000 ml = 0.00059 mol Th⁴⁺

4. Number of moles of F⁻

Use the mole ratio:

       [tex]\dfrac{4molF^-}{1molTh^{4+}}\times 0.00059molTh^{4+}=0.00236molF^-[/tex]

5. Number of moles of HF

Assuming 100% dissociation 1 mol of F⁻  is equal to 1 mol of HF

6. Convert 1 mol of HF to mass in grams

  • molar mass of HF: 20.0063 g/mol
  • mass = molar mass × number of moles
  • mass = 20.0063 g/mol × 0.00236 mol = 0.0472 g

7. Calculate 50% excess:

Multiply by 1.5

  • 0.0472g × 1.5 = 0.0708 g

8. Use the percent concentration

% = ( mass of solute / mass of solution ) × 100

  • 0.491/100 = 0.0708g / mass of solution
  • mass of solution = 0.708 g / 0.00491) = 14.4 g