A mixture of sodium hydroxide and strontium hydroxide contains a total of 0.19 moles of the two compounds. It requires 100.0 mL

Question

3 years ago

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of 2.969 M HCl to neutralize all of the base. How many moles of sodium hydroxide were in the original mixture

1 answer:

9966 [12] · Answer 1 · 2022-06-02T10:00:48+03:00

Answer:

0.0831 moles of NaOH were in the original mixture

Explanation:

The strontium hydroxide, Sr(OH)2, requires 2 moles of HCl per mole of hydroxide to be neutralized.

The sodium hydroxide, NaOH, requires 1 mol of HCl per mol of hydroxide.

To solve this question we must write 2 equations:

(1) 0.19mol = X + Y

Where X = Moles NaOH; Y = Moles Sr(OH)2

The moles of HCl required are:

0.100L * (2.969mol / L) = 0.2969 moles HCl

0.2969 mol = X + 2Y(2)

Replacing (1) in (2):

0.19mol - Y = X

0.2969 mol = (0.19mol - Y) + 2Y

0.1069mol = Y

Moles X = Moles NaOH:

0.19mol = X + 0.1069mol

X = 0.0831 moles of NaOH were in the original mixture