Answer:
Empirical formula is CH₄
Molecular formula = C₂H₈
Explanation:
Mass of carbon = 37.5 g
Mass of hydrogen = 12.5 g
Molecular weight = 32 g/mol
Molecular formula = ?
Empirical formula = ?
Solution:
Number of gram atoms of C = 37.5 g /12g/mol = 3.125
Number of gram atoms of H = 12.5 g / 1.008 g/mol= 12.4
Atomic ratio:
C : H
3.125/3.125 : 12.4 /3.125
1 : 4
C : H : = 1 : 4
Empirical formula is CH₄
Molecular formula:
Molecular formula = n (empirical formula)
n = molar mass of compound / empirical formula mass
n = 32 / 16
n = 2
Molecular formula = n (empirical formula)
Molecular formula = 2 ( CH₄)
Molecular formula = C₂H₈
Answer:
3. 75.0%
Explanation:
2 ClO2(g) + F2(g) → 2 FClO2(g)
First order with respect to ClO2 and F2.
This means the rate equation is given as;
Rate = k [ClO2][F2]
When the initial concentrations of ClO2 and F2 are equal?
Let's assume an initial value of 1 for both reactants, so rate equation is given as;
Rate = k * 1 * 1 = k
The rate after 25% of the F2 has reacted is what percent of the initial rate?
The concentration left of F2 is 75% ( 100% - 25%) = 0.75
Concentration of ClO2 remains 1.
So rate equation is given as;
Rate = k * 1 * 0.75 = 0.75 k
Comparing 0.75k and k.
This means our answer is;
3. 75.0%
First one??
I believe this is the correct answer
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.
