NH_3 and the other containing HCI, at 25 degree C are connected by a a tube that
ID: 1056666 • Letter: N
Question
NH_3 and the other containing HCI, at 25 degree C are connected by a a tube that is 1.00 times 10^2 cm long, which has a volume of 325 cm^3, where along this tube will the two gases meet and react? Each flask contains 25.5 g of reactant and the gases react until one is completely consumed. How much NH_4CI will be made? Which gas will be in excess and what will be the final pressure of the system after the reaction is complete? Assume the volume of the NH_4CI is negligible. R = 0.0821 L-atm/mol-KExplanation / Answer
The gases reacts in the tube toward HCl side of tube.As HCl is heavy than NH3, moves slow than ammonia.
2) No. of moles of NH3 = weight in g / Mol wt
= 25.5 / 17
= 1.5 moles
No. of moles of HCl = weight in g / Mol wt
= 25.5 / 36.5
= 0.698 moles
that is moles of ammonia is greater than moles of HCl.
0.698 moles of HCl required to react 0.698 moles of ammonia.
Moles of ammonia remaining = 1.5 - 0.698 = 0.802 moles
Therefore ammnia gas is in excess.
HCl is limiting reagent.
36.5 g of HCl gives 53.5 g of NH4Cl
25.5 g HCl gives = 25.5 x 53.5 / 36.5
= 1364.25 / 36.5
= 37.37 g of NH4Cl is formed
Pressure =
PV = nRT
P = nRT / V
= 0.802 x 0.0824 x 398 [ 25 + 273 = 298 k ]
= 26.301 L
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