Answer:
A) 31.22
Explanation:
The reaction of sulfuric acid with NaOH is:
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2H₂O
To solve this problem we need to determine the moles of acid that will react, and, using the chemical equation we can determine the moles of NaOH and the volume that a 0.2389M NaOH solution would require to neutralize it.
<em>Moles H₂SO₄ (Molar mass: 98.08g/mol):</em>
0.9368g * 39.04% = 0.3657g H₂SO₄ * (1mol / 98.08g) =
3.7289x10⁻³moles H₂SO₄
And moles of NaOH that you require to neutralize the acid are:
3.7289x10⁻³moles H₂SO₄ * (2 moles NaOH / 1 mole H₂SO₄) =
7.4578x10⁻³ moles NaOH
Using a 0.2389M NaOH solution:
7.4578x10⁻³ moles NaOH * (1L / 0.2389mol) = 0.03122L = 31.22mL
Right answer is:
<h3>A) 31.22
</h3>
Answer:
67.1%
Explanation:
Based on the chemical equation, if we determine the moles of sodium carbonate, we can find the moles of NaHCO₃ that reacted and its mass, thus:
<em>Moles Na₂CO₃ - 105.99g/mol-:</em>
6.35g * (1mol / 105.99g) = 0.0599 moles of Na₂CO₃ are produced.
As 1 mole of sodium carbonate is produced when 2 moles of NaHCO₃ reacted, moles of NaHCO₃ that reacted are:
0.0599 moles of Na₂CO₃ * (2 moles NaHCO₃ / 1 mole Na₂CO₃) = 0.1198 moles of NaHCO₃
And the mass of NaHCO₃ in the sample (Molar mass: 84g/mol):
0.1198 moles of NaHCO₃ * (84g / mol) = 10.06g of NaHCO₃ were in the original sample.
And percent of NaHCO₃ in the sample is:
10.06g NaHCO₃ / 15g Sample * 100 =
<h3>67.1%</h3>
