Answer :
Answer:
C) 0.027
Explanation:
In this case we can start with the reaction between [tex]NaOH[/tex] and [tex]H_2SeO_4[/tex], so:
[tex]H_2SeO_4~+~NaOH~->~Na_2SeO_4~+~H_2O[/tex]
We have an acid ([tex]H_2SeO_4[/tex]) and a base ([tex]NaOH[/tex]), therefore we will have an acid-base reaction in which a salt is produced ([tex]Na_2SeO_4[/tex]) and water ([tex]H_2O[/tex]).
Now we can balance the reaction:
[tex]H_2SeO_4~+~2NaOH~->~Na_2SeO_4~+~2H_2O[/tex]
If we have the volume (45 mL= 0.045 L) and the concentration (0.3 M) of the acid we can calculate the moles using the molarity equation:
[tex]M=\frac{mol}{L}[/tex]
[tex]0.3~M~=~\frac{mol}{0.045~L}[/tex]
[tex]mol=0.3~M*0.045~L=0.0135~mol~of~H_2SeO_4[/tex]
In the balanced reaction, we have a 2:1 molar ratio between the acid and the base (for each mol of [tex]H_2SeO_4[/tex] 2 moles of [tex]NaOH[/tex] are consumed), with this in mind we can calculate the moles of NaOH:
[tex]0.0135~mol~of~H_2SeO_4\frac{2~mol~NaOH}{1~mol~of~H_2SeO_4}=0.027~mol~NaOH[/tex]
I hope it helps!