Answer: 0.1161 grams of mercury(II) sulfide) form.
Explanation:
To calculate the number of moles for given molarity, we use the equation:
.....(1)
a) Molarity of
solution = 0.10 M
Volume of solution = 0.020 L
Putting values in equation 1, we get:


b) Molarity of
solution = 0.010 M
Volume of solution = 0.050 L
Putting values in equation 1, we get:


According to stoichiometry :
1 mole of
reacts with 1 mole of 
Thus 0.0005 moles of
reacts with=
moles of 
Thus
is the limiting reagent and
is the excess reagent.
According to stoichiometry :
1 mole of
forms= 1 mole of 
Thus 0.0005 moles of
forms=
moles of 
mass of 
Thus 0.1161 grams of mercury(II) sulfide) form.