If 20.0grams KOH reacts with 15.0grams(NH4)2So4 according to the reaction bewlow calculate the iters of NH3..?

produced at 500 torr and 20 degrees celcuis. 2KOH+(NH4)2SO4->K2SO4+2NH3+2H2... balanced)

the answer is
a. 13.0L
b.8.31L
c.21.3L
d.3.12L

Answer:
Molar mass KOH = 56.1 g/mol
20.0 / 56.1 = 0.356 moles KOH

Molar mass (NH4)2SO4 = 132 g/mol
15.0 / 132 = 0.114 moles (NH4)2SO4

The ratio between KOH and (NH4)2SO4 is 2 : 1
2 : 1 = x : 0.114
x = 0.228 = moles KOH needed for the reaction
We have 0.356 moles KOH so it is in excess and (NH4)2SO4 is the limiting reactant

The ratio between (NH4)2SO4 (limiting reactant) and NH3 is 1 : 2
1 : 2 = 0.114 : x
x = 0.228 = moles NH3 produced

p = 500 / 760 = 0.658 atm
T = 20 + 273 = 293 K

V = nRT / p = 0.228 x 0.0821 x 293 / 0.658 = 8.32 L of NH3

the answer is b

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