6.The original Listerine formula contains 26.9% alcohol (v/v). How many liters of Listerine can be prepared from 4.5 L of pure a
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Answer is in picture below.
Use 100 grams of the compound:
ω(Cl) = 85.5% ÷ 100%.
ω(Cl) = 0.855; mass percentage of the chlorine in the compound.
m(Cl) = 0.855 · 100 g.
m(Cl) = 85.5 g; mass of chlorine.
m(C) = 100 g - 85.5 g.
m(C) = 14.5 g; mass of carbon.
n(Cl) = m(Cl) ÷ M(Cl).
n(Cl) = 85.5 g ÷ 35.45 g/mol.
n(Cl) = 2.41 mol; amount of chlorine.
n(C) = 14.5 g ÷ 12 g/mol.
n(C) = 1.21 mol; amount of carbon.
n(Cl) : n(C) = 2.41 mol : 1.21 mol = 2 : 1.
This compound is dichlorocarbene CCl₂.
