Question

Consider a tube that is 1.0 meter long. If NH3 gas is introduce at the 0-m mark at the same time HCl gas is introduced at the 1-m mark, how far down the tube will the ring of gas occur?

Answer 1

|---------------------|
0...x.........|..1-x..|<< 1meter

distance NH3 travels = x
distance HCl travels = 1-x
(x/1-x) = sqrt(molar mass HCl/molar mass NH3)
Solve for x and 1-x.
I estimated x = about 60 cm and 1-x about 40 cm.

Answer 2

thank you