ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J.
1 answer:
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<span>4NH</span>₃<span> + 6NO → 5N</span>₂<span> + 6H</span>₂<span>O
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer
mol of NO = </span>
=
= 0.93 mol
Based on the balance equation mole ratio of NH₃ : NO is 4 : 6
= 2 : 3
If mol of NO = 0.93 mol
then mol of NH₃ =
= 0.62 mol
Mass of ammonia = mol × molar mass
= 0.62 mol × 17.03 g/mol
= 10.54 g
Therefore B is the best answer