Answer:
101 L
Explanation:
35.0 g KOH ÷ 56.09 g/mol KOH × (1 mol H2O/ 1 mol KOH) × 18 g/mol H2O = 11.2 g H2O
35.0 g HCl ÷ 36.45 g/mol HCl × (1 mol H2O/ 1 mol HCl) × 18 g/mol H2O = 17.3 g H2O
35.0 g KOH is the limiting reactant
the reaction is
2NO(g) + 2H2(g) <—> N2(g) + 2H2O (g)
Kc = [N2] [ H2O]^2 / [NO]^2 [ H2]^2
Given
moles of NO = 0.124 therefore [NO] = moles /volume = 0.124 /2 = 0.062
moles of H2 = 0.0240 , therefore [H2] = moles / volume = 0.0240 / 2 = 0.012
moles of N2 = 0.0380 , therefore [N2] = moles / volume = 0.0380 / 2 = 0.019
moles of H2O = 0.0276 , therefore [H2O] = moles / volume = 0.0276 / 2 = 0.0138
Kc = (0.019) ( 0.0138)^2 / (0.062)^2 ( 0.012)^2 = 6.54