Respuesta :
Answer: 100 ml
Explanation:
1) Convert 1.38 g of Fe₂S₃ into number of moles, n
i) Formula: n = mass in grass / molar mass
ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃
2) Use the percent yield to calculate the theoretical amount:
65% = 0.65 = actual yield/ theoretical yield =>
theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃
3) Chemical equation:
3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
4) Stoichiometrical mole ratios:
3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl
5) Proportionality:
2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
=> x = 0.020 mol FeCl₃
6) convert 0.020 mol to volume
i) Molarity formula: M = n / V
ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
Explanation:
1) Convert 1.38 g of Fe₂S₃ into number of moles, n
i) Formula: n = mass in grass / molar mass
ii) molar mass of Fe₂S₃ =2 x 55.8 g/mol + 3 x 32.1 g/mol = 207.9 g/mol
iii) n = 1.38 g / 207.9 g/mol = 0.00664 moles of Fe₂S₃
2) Use the percent yield to calculate the theoretical amount:
65% = 0.65 = actual yield/ theoretical yield =>
theoretical yield = actual yield / 0.65 = 0.00664 moles / 0.65 = 0.010 mol Fe₂S₃
3) Chemical equation:
3 Na₂S(aq) + 2 FeCl₃(aq) → Fe₂S₃(s) + 6 NaCl(aq)
4) Stoichiometrical mole ratios:
3 mol Na₂S : 2 mol FeCl₃ : 1 mol Fe₂S₃ : 6 mol NaCl
5) Proportionality:
2moles FeCl₃ / 1 mol Fe₂S₃ = x / 0.010 mol Fe₂S₃
=> x = 0.020 mol FeCl₃
6) convert 0.020 mol to volume
i) Molarity formula: M = n / V
ii) V = n / M = 0.020 mol / 0.2 M = 0.1 liter = 100 ml
100mL of [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to react with an excess of [tex]\rm \bold {Na_2 S}[/tex] to produce 1.38 g of [tex]\rm \bold { Fe_2S_3}[/tex] .
The reaction is
[tex]\rm \bold{3 Na_2S(aq) + 2 FeCl_2(aq) \rightarrow Fe_2S_3(s) + 6 NaCl(aq)}[/tex]
The mass of [tex]\rm \bold { Fe_2S_3}[/tex] is 1.38 g
The number of moles of [tex]\rm \bold { Fe_2S_3}[/tex] will be 0.00664 moles.
If the if the percent yield for the reaction is 65.0% or 0.65.
[tex]\rm \bold { Theoritical Yield = \frac{0.00664}{0.65} }\\\\\rm \bold { Theoritical Yield = 0.01 Moles}[/tex]
Since 2 mole of [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to produce 1 mole of [tex]\rm \bold { Fe_2S_3}[/tex].
[tex]\rm \bold { Fe_2Cl_3}[/tex] concentration is 0.020 mol
To find the from molarity formula
[tex]\rm \bold { V = \frac{n}{M} = \frac{0.020 mol}{ 0.2 M} / = 0.1 liter = 100 ml}\\\\\rm \bold { V = \frac{0.020 mol}{ 0.2 M} }\\\\\rm \bold { V = 0.1 liter = 100 ml}[/tex]
Hence we can conclude that 100mL of [tex]\rm \bold { Fe_2Cl_3}[/tex] is needed to react with an excess of [tex]\rm \bold {Na_2 S}[/tex] to produce 1.38 g of [tex]\rm \bold { Fe_2S_3}[/tex] .
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