Respuesta :
Answer:
1.82 g is the maximum mass of CuS.
Explanation:
Considering:
[tex]Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}[/tex]
Or,
[tex]Moles =Molarity \times {Volume\ of\ the\ solution}[/tex]
Given :
For [tex]CuCl_2[/tex] :
Molarity = 0.500 M
Volume = 38.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 38.0×10⁻³ L
Thus, moles of [tex]CuCl_2[/tex] :
[tex]Moles=0.500 \times {38.0\times 10^{-3}}\ moles[/tex]
Moles of [tex]CuCl_2[/tex] = 0.019 moles
For [tex](NH_4)_2S[/tex] :
Molarity = 0.600 M
Volume = 42.0 mL
The conversion of mL to L is shown below:
1 mL = 10⁻³ L
Thus, volume = 42.0×10⁻³ L
Thus, moles of [tex](NH_4)_2S[/tex] :
[tex]Moles=0.600 \times {42.0\times 10^{-3}}\ moles[/tex]
Moles of [tex](NH_4)_2S[/tex] = 0.0252 moles
According to the given reaction:
[tex]CuCl_2_{(aq)}+(NH_4)_2S_{(aq)}\rightarrow CuS_{(s)}+2NH_4Cl_{(aq)}[/tex]
1 mole of [tex]CuCl_2[/tex] reacts with 1 mole of [tex](NH_4)_2S[/tex]
So,
0.019 mole of [tex]CuCl_2[/tex] reacts with 0.019 mole of [tex](NH_4)_2S[/tex]
Moles of [tex](NH_4)_2S[/tex] = 0.019 mole
Available moles of [tex](NH_4)_2S[/tex] = 0.0252 mole
Limiting reagent is the one which is present in small amount. Thus, [tex]CuCl_2[/tex] is limiting reagent.
The formation of the product is governed by the limiting reagent. So,
1 mole of [tex]CuCl_2[/tex] gives 1 mole of [tex]CuS[/tex]
0.019 mole of [tex]CuCl_2[/tex] gives 0.019 mole of [tex]CuS[/tex]
Moles of [tex]CuS[/tex] formed = 0.019 moles
Molar mass of [tex]CuS[/tex] = 95.611 g/mol
Mass of [tex]CuS[/tex] = Moles × Molar mass = 0.019 × 95.611 g = 1.82 g
1.82 g is the maximum mass of CuS.
