Answer:
3 moles of CaO
Explanation:
The chemical balanced equation is;
2Ca + O₂ → 2CaO
The ratio of calcium(Ca) to Calcium Oxide(Cao) in this equation is 2:2.
Now, when 3 moles of calcium react with oxygen, Cao produced will be;
3 moles of O₂ × 2 moles of CaO/2 moles of Cao = 3 moles of CaO
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.
1 mole = 6.022*10^23 atoms
4.2 moles sulfur = 4.2* 6.022 * 10^23 sulfur atoms = 2.421*10^24 sulfur atoms
1 mole sulfur = <span>32.065 g sulfur
4.2 moles sulfur = 4.2 * </span>32.065 g sulfur = 128.901 g sulfur
