Answer:
Molar percent of sodium in original mixture is 88,50%
Explanation:
The last reaction is:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
The moles of BaClâ‚‚ are:
0,132L Ă— 3,80M = 0,502 moles of BaClâ‚‚
As the amount of BaClâ‚‚ is the maximum possible to produce BaSOâ‚„, the moles of BaClâ‚‚ must be the same than moles of Naâ‚‚SOâ‚„.
0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄
These moles of Naâ‚‚SOâ‚„ comes from:
2 Na + H₂SO₄ → Na₂SO₄ + H₂
As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Naâ‚‚SOâ‚„
0,502 moles of Naâ‚‚SOâ‚„ Ă—[tex]\frac{2molesNa}{1moleNa_{2}SO_{4}}[/tex]Ă— 22,99 g/mole = 23,08 g of Na
Molar percent of sodium in original mixture is:
[tex]\frac{23,08g}{26,08g}*100[/tex] = 88,50%
I hope it helps
