Answer:
5.53 grams is the mass of NaOH in the mixture.
Explanation:
Let the mass of NaOH be x and [tex]Ba(OH)_2[/tex]be y.
x + y = 10.1 g..[1]
Moles of NaOH = [tex]\frac{x}{40 g/mol}[/tex]
Moles of [tex]Ba(OH)_2=\frac{y}{171 g/mol}[/tex]
Moles of HCl = n
Volume of HCl solution = 107.8 mL = 0.1078 L( 1 mL = 0.001 L)
Molarity of HCl solution = 1.53 M
[tex]n=1.53M\times 0.1078 L=0.164934 mol[/tex]
0.38709 moles neutralizes all the NaOH and [tex]Ba(OH)_2[/tex] present in solution.So, This manes that;
[tex]\frac{x}{40 g/mol}+\frac{y}{171 g/mol}=0.164934 mol[/tex]
[tex]171 x+40y=1128.15[/tex] ..[2]
On solving [1] and [2] we get ;
x = 5.53 g
y = 4.57 g
5.53 grams is the mass of NaOH in the mixture.
