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A 0.652-g sample of a pure strontium halide reacts with
Chapter 3, Problem 98P(choose chapter or problem)
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid, and the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. What is the formula of the original halide?
Questions & Answers
QUESTION:
A 0.652-g sample of a pure strontium halide reacts with excess sulfuric acid, and the solid strontium sulfate formed is separated, dried, and found to weigh 0.755 g. What is the formula of the original halide?
ANSWER:Step 1 of 2
Here, we are going to determine the formula of the original halide present in the reaction.
The given equation of the reaction is
Given that,
Mass of = 0.652 g
Mass of = 0.755 g
Therefore,
The moles of = (0.755 g )
= 0.0041 mol
Thus,
0.652 g = 0.0041 mol
Thus, the molar mass of = = 158.630 g/mol