Answer:
The answer to your question is 22.5 moles of H₂SO₄
Explanation:
Data
moles of H₂SO₄ = ?
moles of Al = 15
Balanced chemical reaction
2Al + 3H₂SO₄ ⇒ Al₂(SO₄)₃ + 3H₂
To solve this problem use proportions and cross multiplication. Use the coefficients of the balanced chemical reaction.
2 moles of Al ------------------- 3 moles of H₂SO₄
15 moles of Al ------------------ x
x = (15 x 3) / 2
x = 45/2
x = 22.5 moles of H₂SO₄
-Conclusion
22.5 moles of H₂SO₄ react completely with 15 moles of Al.
Answer:
ΔH for the the reaction NO(g) + O(g) ⇒ NO₂(g) is ΔH= -304.1 kJ
Explanation:
<u>The complete question is:</u>
Consider the chemical equations shown here.
NO(g) + O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH= -198.9 kJ
)
1.5 O₂(g) ⇒ O₃(g) (ΔH= 142.3 kJ
)
O(g) ⇒ 0.5 O₂(g) (ΔH= -247.5 kJ)
What is ΔH for the reaction shown below?
NO(g) + O(g) ⇒ NO₂(g)
Solution:
We have to use the Hess's Law: if a series of reagents react to give a series of products, the heat of reaction released or absorbed is independent of whether the reaction is carried out in one, two or more stages. That means enthalpy changes are additive.
NO(g)+ O₃(g) ⇒ NO₂(g) + O₂(g) (ΔH₁= -198.9 kJ
)
+
1.5 O₂(g) ⇒ O₃(g) (ΔH₂= 142.3 kJ
)
+
O(g) ⇒ 0.5 O₂(g) (ΔH₃= -247.5 kJ)
=
NO(g) + O₃(g) + 1.5 O₂(g) + O(g) ⇒ NO₂(g) + O₂(g) + O₃(g) + 0.5 O₂(g)
We remove the compounds that are in both members of the reaction:
NO(g) + O(g) ⇒ NO₂(g)
We only have to add the reactions so we add the value of each enthalpy change.
ΔH for the the reaction is given by:
ΔH= ΔH₁ + ΔH₂ + ΔH₃= -198.9 kJ +142.3 kJ -247.5 kJ= -304.1 kJ
Answer:
Radioisotopes are radioactive isotopes of an element. They can also be defined as atoms that contain an unstable combination of neutrons and protons, or excess energy in their nucleus.
Explanation:
Answer:
As the molecules of a liquid are cooled they slow down. As the molecules slow down they take up less volume. Taking up less room because of the molecules lower energy causes the liquid to contract.
Explanation:
Answer:
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant. Explanation:
Moles of calcium nitrate =
Moles of ammonium fluoride =
According to reaction , 2 moles of ammonium fluoride reacts with 1 mole of calcium nitrate.
Then 1.046 moles of ammonium fluoride will react with :
calcium nitarte .
This means that ammonium fluoride is in excess amount and calcium nitrate is in limiting amount.
Hence, calcium nitrate is a limiting reactant.
So, amount of dinitrogen monoxide will depend upon moles of calcium nitrate.
So, according to reaction , 1 mole of calcium nitarte gives 2 moles of dinitrogen monoxide gas .
Then 0.1908 moles of calcium nitrate will give:
of dinitrogen monoxide gas.
Mass of 0.03816 moles of dinitrogen monoxide gas:
0.03816 mol × 44 g/mol = 16.79 g
16.79 grams of dinitrogen monoxide are present after 31.3 g of calcium nitrate and 38.7 g of ammonium fluoride react completely. And calcium nitrate is a limiting reactant.