The experimental group is the group that is left alone during the experiment
1 answer:
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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.