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Question

The volume of HCl(g) at STP required to clean up all NH3 molecules adsorbed on surface is:

A
0.175 L
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B
0.238 L
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C
0.432 L
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D
0.526 L
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Solution

The correct option is D 0.526 L
Total moles of NH3 adsorbed on surface =1.415×10226.023×1023=0.0235.
Moles of NH3 before adsorption =100×21000=0.2.
So, moles of NH3 left after adosrption = moles of HCl required =0.20.0235=0.175.
Also, moles of HCl required to clean surface = moles of NH3 adsorbed.
nHCl=0.0235
1×V=0.0235×0.0821×273
VHCl(g)=0.526 L

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