How many moles of cesium xenon heptafloride can be produced from the reaction 12 mol cesium Floride with 14 mol xenon hexaflorid
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Answer:
The molarity of this solution is 0,09254M
Explanation:
The concentration of the Ni²⁺ solution is:
Ni²⁺ + EDTA⁴⁻ → Ni(EDTA)²⁻
0,03935L × 0,01307M = 5,143x10⁻⁴ moles Ni²⁺ ÷ 0,03010L =<em>0,01709M Ni²⁺</em>
25,00 mL of this solution contain:
0,01709M × 0,02500L = 4,2716x10⁻⁴ moles of Ni²⁺
The moles of Ni²⁺ that are in excess and react with EDTA⁴⁻ are:
0,01015L × 0,01307M = 1,3266x10⁻⁴ moles of Ni²⁺
Thus, moles of Ni²⁺ that react with CN⁻ are:
4,2716x10⁻⁴ - 1,3266x10⁻⁴ = 2,9450x10⁻⁴ moles of Ni²⁺
For the reaction:
4CN⁻ + Ni²⁺ → Ni(CN)₄²⁻
Four moles of CN⁻ react with 1 mole of Ni²⁺:
2,9450x10⁻⁴ moles of Ni²⁺ × = <em>1,178x10⁻³ moles of CN⁻</em>
As the volume of cyanide solution is 12,73mL. The molarity of this solution is:
<em>1,178x10⁻³ moles of CN⁻ ÷ 0,01273L = </em><em>0,09254M</em>
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