Respuesta :
Answer: 0.1161 grams of mercury(II) sulfide) form.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in L)}}[/tex] .....(1)
a) Molarity of [tex]Na_2S[/tex] solution = 0.10 M
Volume of solution = 0.020 L
Putting values in equation 1, we get:
[tex]0.10M=\frac{\text{Moles of }Na_2S}{0.020L}\\\\\text{Moles of Na_2S}={0.10mol/L\times 0.020}=0.002mol[/tex]
[tex]\text {Moles of}Na_2S=0.10M\times 0.020L=0.002mol[/tex]
b) Molarity of [tex]Hg(NO_3)_2[/tex] solution = 0.010 M
Volume of solution = 0.050 L
Putting values in equation 1, we get:
[tex]0.010M=\frac{\text{Moles of }Hg(NO_3)_2}{0.050L}\\\\\text{Moles of }Hg(NO_3)_2={0.010mol/L\times 0.050}=0.0005mol[/tex]
[tex]Na_2S+Hg(NO_3)_2\rightarrow HgS+2NaNO_3[/tex]
According to stoichiometry :
1 mole of [tex]Hg(NO_3)_2[/tex] reacts with 1 mole of [tex]Na_2S[/tex]
Thus 0.0005 moles of [tex]HgNO_3[/tex] reacts with=[tex]\frac{1}{1}\times 0.0005=0.0005[/tex] moles of [tex]Hg(NO_3)_2[/tex]
Thus [tex]Hg(NO_3)_2[/tex] is the limiting reagent and [tex]Na_2S[/tex] is the excess reagent.
According to stoichiometry :
1 mole of [tex]Hg(NO_3)_2[/tex] forms= 1 mole of [tex]Hg_2S[/tex]
Thus 0.0005 moles of [tex]Hg(NO_3)_2[/tex] forms=[tex]\frac{1}{1}\times 0.0005=0.0005[/tex] moles of [tex]Hg_2S[/tex]
mass of [tex]H_2S=moles\times {\text {Molar mass}}=0.0005mol\times 232.2g/mol=0.1161g[/tex]
Thus 0.1161 grams of mercury(II) sulfide) form.
