A .5121 g sample of a pure soluble chloride compound is dissolved in water, and
ID: 786697 • Letter: A
Question
A .5121 g sample of a pure soluble chloride compound is dissolved in water, and all of the chloride ion is precipitated as AgCl by the addition of an excess of silver nitrate. The mass of the resulting AgCl is found to be 1.1942 g.
What is the mass percentage of chlorine in the original compound?
A student determines the iron(III) content of a solution by first precipitating it as iron(III) hydroxide, and then decomposing the hydroxide to iron(III) oxide by heating. How many grams of iron(III) oxide should the student obtain if his solution contains 59.0 mL of 0.417 M iron(III) nitrate?
Explanation / Answer
(1) Moles of Cl = moles of AgCl = mass/molar mass of AgCl
= 1.1942/143.3214 = 0.00833232 mol
Mass of Cl = moles x molar mass of Cl
= 0.00833232 x 35.4532 = 0.295407 g
Mass% of Cl = mass of Cl/mass of sample x 100%
= 0.295407/0.5121 x 100%
= 57.69%
(2) 2 Fe(NO3)3 => Fe2O3
Moles of Fe(NO3)3 = volume x concentration of Fe(NO3)3
= 59.0/1000 x 0.417 = 0.024603 mol
Moles of Fe2O3 = 1/2 x moles of Fe(NO3)3
= 1/2 x 0.024603 = 0.0123015 mol
Mass of Fe2O3 = moles x molar mass of Fe2O3
= 0.0123015 x 159.69
= 1.964 g = 1.96 g
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