Answer:
The answer would be at 120°C
The volume of H₂ : = 15.2208 L
<h3>Further explanation</h3>
Given
Reaction
2 As (s) + 6 NaOH (aq) → 2 Na₃AsO₃ (s) + 3 H₂ (g)
34.0g of As
Required
The volume of H₂ at STP
Solution
mol As (Ar = 75 g/mol) :
= mass : Ar
= 34 g : 75 g/mol
= 0.453 mol
From the equation, mol ratio As : H₂ = 2 : 3, so mol H₂ :
=3/2 x mol As
=3/2 x 0.453
= 0.6795
At STP, 1 mol = 22.4 L, so :
= 0.6795 x 22.4 L
= 15.2208 L
Answer:
K = 3.37
Explanation:
2 NH₃(g) → N₂(g) + 3H₂(g)
Initially we have 4 mol of ammonia, and in equilibrium we have 2 moles, so we have to think, that 2 moles have been reacted (4-2).
2 NH₃(g) → N₂(g) + 3H₂(g)
Initally 4moles - -
React 2moles 2m + 3m
Eq 2 moles 2m 3m
We had produced 2 moles of nitrogen and 3 mol of H₂ (ratio is 2:3)
The expression for K is: ( [H₂]³ . [N₂] ) / [NH₃]²
We have to divide the concentration /2L, cause we need MOLARITY to calculate K (mol/L)
K = ( (2m/2L) . (3m/2L)³ ) / (2m/2L)²
K = 27/8 / 1 → 3.37
