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Mumz [18]
2 years ago
12

If there are 40 mol of NBr3 and 48 mol of NaOH , what is the excess reactant ?

Chemistry
2 answers:
tekilochka [14]2 years ago
5 0
2NBr₃ + 3NaOH = N₂ + 3HOBr + 3NaBr
40 mol    48 mol

NBr₃:NaOH = 2:3

40:48 = 2:2.4 = 2.5:3

NBr₃ is the excess reactant
Paul [167]2 years ago
3 0

Answer:

NBr3

Explanation:

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1) Assume a general equation for the ionization of the weak acid:

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HA ⇄ H⁺ + A⁻

2) Then, the expression for the ionization constant is:

Ka = [H⁺][A⁻] / [HA]

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3) So, you need to determine [H⁺] which you do from the pH.

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