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Nadya [2.5K]
4 years ago
14

A solution contains cr3+ ions. the addition of 0.063 l of 1.50 m naf solution was needed to completely precipitate the chromium

ions as crf3(s). what was the mass of cr3+ ions in the original solution?
Chemistry
1 answer:
steposvetlana [31]4 years ago
6 0

Cr{3+} + 3 NaF → CrF3 + 3 Na{+} <span>

First calculate the total mols of NaF. 

(0.063 L) x (1.50 mol/L NaF) = 0.0945 mol NaF total </span>

 

Using stoichiometric ratio:

<span>0.0945 mol NaF * (1 mol Cr3+ / 3 mol NaF) * (51.9961 g Cr3+/mol) = 1.6379 g Cr3+</span>
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