2. How many grams of N2 are required to produce 10.0 grams of NH3? N2 + 3 H2 - 2 NH3
1 answer:
8.24g
Explanation:
Given parameters;
Mass of NH₃ = 10g
Unknown:
Mass of N₂ = ?
Solution:
Reaction equation:
N₂ + 3H₂ → 2NH₃
Step 1: Convert 10.0 g of NH₃ into moles of NH₃
number of moles =
molar mass of NH₃ = 14 + 3(1) = 17g/mol
Number of moles = = 0.59mole
Step 2: Convert the moles of NH₃ calculated in Step 1 into moles of N₂
1 mole of N₂ produced 2 mole of NH₃
0.59 moles of ammonia would be formed by mole = 0.29mole of N₂
Step 3: Convert the moles of N₂ calculated in Step 2 into grams of N₂.
Mass of N₂ = number of moles x molar mass
Molar mass of N₂ = 14 x 2 = 28g/mol
Mass of N₂ = 0.29 x 28 = 8.24g
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Explanation:
Relation between entropy change and specific heat is as follows.
The given data is as follows.
mass = 500 g, = 24.4 J/mol K
= 500 K,
= 250 K
Mass number of copper = 63.54 g /mol
Number of moles =
=
= 7.86 moles
Now, equating the entropy change for both the substances as follows.
=
= 750
So, =
- For the metal block A, change in entropy is as follows.
=
= -3.04 J/ K mol
- For the block B, change in entropy is as follows.
=
= 4.296 J/Kmol
And, total entropy change will be as follows.
= 4.296 + (-3.04)
= 1.256 J/Kmol
Thus, we can conclude that change in entropy of block A is -3.04 J/ K mol and change in entropy of block B is 4.296 J/Kmol.