Answer:
0.478 M
Explanation:
Let's consider the neutralization reaction between KOH and H₂SO₄.
2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O
12.7 mL of 1.50 M H₂SO₄ react. The reacting moles of H₂SO₄ are:
0.0127 L × 1.50 mol/L = 0.0191 mol
The molar ratio of KOH to H₂SO₄ is 2:1. The reacting moles of KOH are 2 × 0.0191 mol = 0.0382 mol
0.0382 moles of KOH are in 80.0 mL. The molarity of KOH is:
M = 0.0382 mol/0.0800 L = 0.478 M
The answer is 57.14%.
First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.
Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g
Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g
mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%
Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>