chem problem?

1)
Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Molar mass of Ba(OH)₂ = (137.3 + 16.0×2 + 1.0×2) g/mol = 171.3 g/mol

Let m g be the mass NaOH in the mixture.
Then, mass of Ba(OH)₂ in the mixture = (10.7 - m) g

NaOH + HCl → NaCl + H₂O
Mole ratio NaOH : HCl = 1 : 1
No. of moles of NaOH reacted = (m g) / (40.0 g/mol) = m/40 mol
No. of moles of HCl needed to react with NaOH = m/40 mol

Ba(OH)₂ + 2HCl → BaCl₂ + 2H₂O
Mole ratio Ba(OH)₂ : HCl = 1 : 2
No. of moles of Ba(OH)₂ reacted = [(10.7 - m) g] / (171.3 g/mol) = (10.7 - m)/171.3 mol
No. of moles of HCl needed to react with Ba(OH)₂ = 2(10.7 - m)/171.3 mol

Total number of moles of HCl needed :
{(m/40) + [(10.7 - m)/171.3]} mol = (1.54 mol/L) × (108.6/1000 L)
(m/40) + [2(10.7 - m)/171.3] = 0.16724
{(m/40) + [2(10.7 - m)/171.3]} × 40 × 171.3 = 0.1672 × 40 × 171.3
171.3m + 856 - 80m = 1146
91.3m = 290
m = 3.18 g
Mass of NaOH = 3.18 g


2)
Mass of Ba(OH)₂ = (10.7 - 3.18) g = 7.52 g

Interesting stoichiometry problem....

40.0g/mol ..... 171.3g/mol
?NaOH(s) + ?Ba(OH)2(s) + ?HCl(aq) --> ?NaCl(aq) + ?BaCl2 + ?HOH(l)
x g ................. 10.7-x ......... 108.6mL, 1.54M

0.1086L x (1.54 mol HCl / 1L) = 0.167 mol HCl

Each mole of NaOH requires 1 mol of HCl. Each mole of Ba(OH)2 requires 2 mol HCl.

...1 ..................... 2 ................. 3
x / 40 + 2(10.7-x) / 171.3) = 0.167 mol HCl
1 .... moles of HCl which reacts with NaOH
2 .... moles of HCl which reacts with Ba(OH)2
3 .... moles of HCl available

Use some algebra and solve for x

x = 3.16g ............. of NaOH
10.7-x = 7.54g ..... of Ba(OH)2

Note: The 234 mL of water in which the solids are dissolved is superfluous. You don't need it to solve the problem.