Leftover: approximately 11.73 g of sulfuric acid.
<h3>Explanation</h3>
Which reactant is <em>in excess</em>?
The theoretical yield of water from Al(OH)₃ is lower than that from H₂SO₄. As a result,
- Al(OH)₃ is the limiting reactant.
- H₂SO₄ is in excess.
How many <em>moles</em> of H₂SO₄ is consumed?
Balanced equation:
2 Al(OH)₃ + 3 H₂SO₄ → Al₂(SO₄)₃ + 6 H₂O
Each mole of Al(OH)₃ corresponds to 3/2 moles of H₂SO4. The formula mass of Al(OH)₃ is 78.003 g/mol. There are 15 / 78.003 = 0.19230 moles of Al(OH)₃ in the five grams of Al(OH)₃ available. Al(OH)₃ is in excess, meaning that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
How many <em>grams</em> of H₂SO₄ is consumed?
The molar mass of H₂SO₄ is 98.076 g.mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.076 = 28.289 g.
How many <em>grams</em> of H₂SO₄ is in excess?
40 grams of sulfuric acid H₂SO₄ is available. 28.289 grams is consumed. The remaining 40 - 28.289 = 11.711 g is in excess. That's closest to the first option: 11.73 g of sulfuric acid.
C. Walking on wet floors shows that Friction is undesirable
The answer is B) relief
Relief is the is the difference in elevation between the highest and lowest points in an area.
Balance the reaction first:
3KOH + H3PO4 —> K3PO4 + 3H2O
So for every mol of H3PO4, you need 3 mol of OH- to fully neutralize the acid, since H3PO4 is polyprotic.
0.0200 L KOH • (2.000 mol KOH / L KOH) • (1 mol H3PO4 / 3 mol KOH) = 0.0133 mol H3PO4
Divide this by the volume of H3PO4 to get the concentration.
0.0133 mol H3PO4 / 0.0250 L = 0.532 M H3PO4
