Answer:
ΔHrxn = -635.14kJ/mol
Explanation:
We can make algebraic operations of reactions until obtain the desire reaction and, ΔH of the reaction must be operated in the same way to obtain the ΔH of the desire reaction (Hess's law). Using the reactions:
(1)Ca(s) + 2 H+(aq) → Ca2+(aq) + H2(g) ΔH = 1925.9 kJ/mol
(2) 2H2(g) + O2 g) → 2 H2O(l) ΔH = −571.68 kJ/mole
(3) CaO(s) + 2 H+(aq) → Ca2+(aq) + H2O(l) ΔH = 2275.2 kJ/mole
Reaction (1) - (3) produce:
Ca(s) + H2O(l) → H2(g) + CaO(s)
ΔH = 1925.9kJ/mol - 2275.2kJ/mol = -349.3kJ/mol
Now this reaction + 1/2(2):
Ca(s) + ½ O2(g) → CaO(s)
ΔH = -349.3kJ/mol + 1/2 (-571.68kJ/mol)
<h3>ΔHrxn = -635.14kJ/mol</h3>
Answer:
The answer to your question is 40 L of NH₃
Explanation:
Data
Volume of NH₃ = x
mass of N₂ = 25 g
mass of H₂ = excess
Balanced chemical reaction
N₂ + 3H₂ ⇒ 2NH₃
Process
1.- Find the molar mass of N₂ and NH₃
N₂ = 14 x 2 = 28g
2NH₃ = 2[ 14 + 3] = 34 g
2.- Write a proportion to solve this problem
28 g of N₂ --------------- 34 g of NH₃
25 g of N₂ ------------- x
x = (25 x 34)/28
x = 30.36 g of NH₃
3.- Calculate the volume of NH₃
17 g of NH₃ -------------- 22.4 L
30.36 g of NH₃ -------- x
x = (30.36 x 22.4) / 17
x = 40 L
15 ppm
1.5 × 10-2 ppm
1.5 × 104 ppm
3.1 ppm
1.5 × 10-2 ppm