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.
Answer:
H2
Explanation:
In the first place it is necessary to consider that these two elements will be their diatomic form initially, that is, H2 and N2.
first we should check the equilibrium constant Kp tables in this case at a temperature of 3000K
Value for dissociation reaction of H2 in Kp = -3.685
Value for dissociation reaction of N2 in Kp = -22.359
Equilibrium constant for H2 dissociation is higher than N2 dissociation. so for this comparation H2 is more likely to dissociate.
Answer:
Mass is the measure of how much matter is in an object
Explanation:
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Answer: i think that it was 800 mg
Explanation:
Answer:
29260J
Explanation:
Given parameters:
Mass of water sample = 100g
Initial temperature = 30°C
Final temperature = 100°C
Unknown:
Energy required for the temperature change = ?
Solution:
The amount of heat required for this temperature change can be derived from the expression below;
H = m c (ΔT)
H is the amount of heat energy
m is the mass
c is the specific heat capacity of water = 4.18J/g°C
ΔT is the change in temperature
Now insert the parameters and solve;
H = 100 x 4.18 x (100 - 30)
H = 100 x 4.18 x 70 = 29260J
