Respuesta :
Answer:
[tex]Pb^{2+}[/tex] is limiting reagent.
Theoretical yield = 262.67 g
% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %
Explanation:
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Given: For [tex]NaCl[/tex]
Given mass = 135.8 g
Molar mass of [tex]NaCl[/tex] = 58.44 g/mol
Moles of [tex]NaCl[/tex] = 135.8 g / 58.44 g/mol = 2.3238 moles
Given: For [tex]Pb^{2+}[/tex]
Given mass = 195.7 g
Molar mass of [tex]Pb^{2+}[/tex] = 207.2 g/mol
Moles of [tex]Pb^{2+}[/tex] = 195.7 g / 207.2 g/mol = 0.9445 moles
According to the given reaction:
[tex]Pb^{2+}_{(aq)}+2NaCl_{(aq)}\rightarrow PbCl_2_{(s)}+2Na^+_{(aq)}[/tex]
1 mole of [tex]Pb^{2+}[/tex] react with 2 moles of [tex]NaCl[/tex]
Also,
0.9445 mole of [tex]Pb^{2+}[/tex] react with 2*0.9445 moles of [tex]NaCl[/tex]
Moles of [tex]NaCl[/tex] = 1.889 moles
Available moles of [tex]NaCl[/tex] = 2.3238 moles (Extra)
Limiting reagent is the one which is present in small amount. Thus, [tex]Pb^{2+}[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of [tex]Pb^{2+}[/tex] gives 1 mole of [tex]PbCl_2[/tex]
0.9445 mole of [tex]Pb^{2+}[/tex] gives 0.9445 mole of [tex]PbCl_2[/tex]
Mole of [tex]PbCl_2[/tex] = 0.9445 moles
Molar mass of [tex]PbCl_2[/tex] = 278.1 g/mol
Mass of [tex]PbCl_2[/tex] = Moles × Molar mass = 0.9445 × 278.1 g = 262.67 g
Theoretical yield = 262.67 g
Given experimental yield = 252.4 g
% yield = (Experimental yield / Theoretical yield) × 100 = (252.4 / 262.67) × 100 = 96.1 %
