Answer to Question #314008 in General Chemistry for Zai

Question

For #1-3, consider the reaction:

3 NaOH(aq) + 3 FeCl3 → NaCl (aq) + 3 Fe(OH)3 (s)

1) What mass of Fe(OH)3 would be produced if 250 mL of 0.250 M NaOH were added to a solution containing excess FeCl3?

2) What volume of 2M NaOH will react with 34.0 g of FeCl3?

3) What is the concentration of 20.50 mL of NaOH if this was neutralized with 20.0 mL of 0.150 M HCl in the reaction

NaOH + HCl → NaCl + HOH

Answer 1

3NaOH_(aq) + 3FeCl₃ → NaCl_(aq) + 3Fe(OH)_3(s)

1) n(NaOH) = 0.250 L * 0.250 mol/L = 0.0625 mol

3 mol NaOH - 3 mol Fe(OH)₃

0.0625 mol NaOH - x mol Fe(OH)₃

x = 0.0625 mol

m(Fe(OH)₃) = n * Mr = 0.0625 mol * 106.87 g/mol = 6.68 g

2) 3 mol * 40.0 g/mol NaOH - 3 mol * 162.2 g/mol FeCl₃

x g NaOH - 34.0 g FeCl₃

x = (34.0 * 3 * 40.0) / (3 * 162.2) = 8.38 g

n(NaOH) = m/Mr = 8.38 g / 40.0 g/mol = 0.2095 mol

V(solution) = n(NaOH)/M(solution) = 0.2095 mol / 2.0 mol/L = 0.10475 L = 104.75 mL

3) NaOH + HCl → NaCl + HOH

n(HCl) = M*V = 0.150 mol/L * 0.020 L = 0.003 mol

1 mol NaOH - 1 mol HCl

x mol NaOH - 0.003 mol HCl

x = 0.003 mol

M(NaOH) = n/V = 0.003 mol / 0.0205 L = 0.146 mol/L