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Answer:
Molar percent of sodium in original mixture is 88,50%
Explanation:
The last reaction is:
BaCl₂ + Na₂SO₄ → BaSO₄ + 2 NaCl
The moles of BaCl₂ are:
0,132L × 3,80M = 0,502 moles of BaCl₂
As the amount of BaCl₂ is the maximum possible to produce BaSO₄, the moles of BaCl₂ must be the same than moles of Na₂SO₄.
0,502 moles of BaCl₂ ≡ 0,502 moles of Na₂SO₄
These moles of Na₂SO₄ comes from:
2 Na + H₂SO₄ → Na₂SO₄ + H₂
As the reaction is in stoichiometric amounts, moles of Na are twice the moles of Na₂SO₄
0,502 moles of Na₂SO₄ ×× 22,99 g/mole = 23,08 g of Na
Molar percent of sodium in original mixture is:
= <em>88,50% </em>
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The volume of oxygen at STP required would be 252.0 mL.
<h3>Stoichiometic problem</h3>The equation for the complete combustion of C2H2 is as below:
The mole ratio of C2H2 to O2 is 2:5.
1 mole of a gas at STP is 22.4 L.
At STP, 100.50 mL of C2H2 will be:
100.50 x 1/22400 = 0.0045 mole
Equivalent mole of O2 according to the balanced equation = 5/2 x 0.0045 = 0.01125 moles
0.01125 moles of O2 at STP = 0.01125 x 22400 = 252.0 mL
Thus, 252.0 mL of O2 gas will be required at STP.
More on stoichiometric problems can be found here: brainly.com/question/14465605
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