Question

How many moles of NaOH is needed to neutralize 45.0ml of 0.30M H2SeO4?

Answer 1

Answer:

0.027 mole of NaOH.

Explanation:

We'll begin by obtaining the number of mole H2SeO4 in 45mL of 0.30M H2SeO4

This is illustrated below:

Molarity of H2SeO4 = 0.3M

Volume of solution = 45mL = 45/1000 = 0.045L

Mole of H2SeO4 =...?

Mole = Molarity x Volume

Mole of H2SeO4 = 0.3 x 0.045

Mole of H2SeO4 = 0.0135 mole

Next, the balanced equation for the reaction. This is given below:

H2SeO4 + 2NaOH –> Na2SeO4 + 2H2O

From the balanced equation above,

1 mole of H2SeO4 required 2 moles of NaOH.

Therefore, 0.0135 mole of H2SeO4 will require = 0.0135 x 2 = 0.027 mole of NaOH.

Therefore, 0.027 mole of NaOH is needed for the reaction.

Answer 2

Answer:

$n_{NaOH}=0.027molNaOH$

Explanation:

Hello,

In this case, the reaction between sodium hydroxide and selenic acid is:

$H_2SeO_4+2NaOH\rightarrow Na_2SeO_4+2H_2O$

Next, we compute the reacting moles of selecnic acid by using its molarity and volume in litres:

$n_{H_2SeO_4}=0.30\frac{mol}{L} *45.0mL*\frac{1L}{1000mL} =0.0135molH_2SeO_4$

Then, since sodium hydroxide and selenic acid are in a 2:1 molar ratio, we compute the moles of sodium hydroxide by stoichiometry:

$n_{NaOH}=0.0135molH_2SeO_4*\frac{2molNaOH}{1molH_2SeO_4*} \\\\n_{NaOH}=0.027molNaOH$

Best regards.