Question

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

Answer 1

`n_(NaOH)=0.027molNaOH`

Step-by-step 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(mol)/(L) *45.0mL*(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*(2molNaOH)/(1molH_2SeO_4*) \\\\n_(NaOH)=0.027molNaOH`

Best regards.

Answer 2

0.027 mole of NaOH.

Step-by-step 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.