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adell [148]
4 years ago
5

How many moles of NaOH is needed to neutralize 45.0 ml of 0.30M H2SeO4? Question 2 options: A) 0.00675 B) 27.0 C) 0.027 D) 0.013

5
Chemistry
1 answer:
Lady bird [3.3K]4 years ago
4 0

Answer:

C) 0.027

Explanation:

In this case we can start with the <u>reaction</u> between NaOH and H_2SeO_4, so:

H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O

We have an <u>acid</u> (H_2SeO_4) and a <u>base</u> (NaOH), therefore we will have an <u>acid-base reaction</u> in which a <u>salt</u> is produced (Na_2SeO_4) and <u>water</u> (H_2O).

Now we can <u>balance the reaction</u>:

H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O

If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can <u>calculate the moles</u> using the molarity equation:

M=\frac{mol}{L}

0.3~M~=~\frac{mol}{0.045~L}

mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4

In the balanced reaction, we have a <u>2:1 molar ratio</u> between the acid and the base (for each mol of H_2SeO_4 2 moles of NaOH are consumed), with this in mind we can calculate the <u>moles of NaOH:</u>

0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH

I hope it helps!

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