The correct answer is approximately 11.73 grams of sulfuric acid.
The theoretical yield of water from Al(OH)3 is lower than that of H₂SO₄. As a consequence, Al(OH)3 is the limiting reactant, H₂SO₄ is in excess.
The balanced equation is:
2Al(OH)₃ + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 6H₂O
Each mole of Al(OH)3 corresponds to 3/2 moles of H₂SO₄. The molecular 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 limiting, which means that all 0.19230 moles will be consumed. Accordingly, 0.19230 × 3/2 = 0.28845 moles of H₂SO₄ will be consumed.
The molar mass of H₂SO₄ is 98.706 g/mol. The mass of 0.28845 moles of H₂SO₄ is 0.28845 × 98.706 = 28.289 g
40 grams of sulfuric acid is available, out of which 28.289 grams is consumed. The remaining 40-28.289 = 11.711 g is in excess, which is closest to the first option, that is, 11.73 grams of H₂SO₄.
Answer: pH = 2,897 , basic![[H+][OH-] = 10^{-14} ==> [H+] = \frac{10^{-14}}{7,89*10^{-12} } =\frac{1}{789} \\pH= -lg([H+]) = 2,897 \\pH basic](https://tex.z-dn.net/?f=%5BH%2B%5D%5BOH-%5D%20%3D%2010%5E%7B-14%7D%20%3D%3D%3E%20%5BH%2B%5D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B7%2C89%2A10%5E%7B-12%7D%20%7D%20%3D%5Cfrac%7B1%7D%7B789%7D%20%5C%5CpH%3D%20-lg%28%5BH%2B%5D%29%20%3D%202%2C897%20%5C%5CpH%3C7%20%3D%3D%3E%20basic)
Explanation:
Answer:
if an atom gains an electron, the ion has negetive charge
Answer:
2 mol H₂O
Explanation:
With the reaction,
- 2H₂(g) + O₂(g) → 2 H₂O(g)
1.55 moles of O₂ would react completely with ( 2*1.55 ) 3.1 moles of H₂. There are not as many moles of H₂, thus H₂ is the limiting reactant.
Now we <u>calculate the moles of H₂O produced</u>, <em>starting from the moles of limiting reactant</em>:
- 2.00 mol H₂ *
= 2 mol H₂O
Answer:
[Br₂] = 1.25M
Explanation:
2NO (g) + Br₂ (g) ⇄ 2NOBr (g)
Eq 0.80M ? 0.80M
That's the situation told, in the statement.
Let's make the expression for Kc
Kc = [NOBr]² / [Br₂] . [NO]²
Kc = 0.80² / [Br₂] . [0.80]²
0.80 = 1 / [Br₂]
[Br₂] = 1 / 0.80 → 1.25
