315791
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How many Liters of I2 are produced when 241.6 grams of KI are reacted with excess CuBr2?

Respuesta :

Answer:

8.15 L I2

Explanation:

So, I have the balanced equation of:  2CuBr2 +4KI  →  2CuI   +   4KBr   + I2

Now, we do:  

241.6g  /  1 mol KI  /  1 mol I2  /  22.4 L

---------------------------------------------------------  =   8.15 L I2

           /  166g KI  /   4 mol KI  /   1 mol I2

Hope this helps! ^u^

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