for each mole of Mg consumed, the rate of H2 is the same. That will be in moles/sec
to change that to volume, use
V=n*24 where n is in moles/sec, so volume will be in liters/sec
Mg + 2HCl -> MgCl2 + H2
Over an interval of 1.00s, the mass of Mg changes by -0.011g
What is the corresponding rate of consumption of HCl (in mol/s)?
I got 9.0E-4 mol/s and that's the right answer. But I'm having problems with the next question.
Calculate the corresponding rate of production of H2 in L/s at 20°C and 101 kPa. Tip: 1.00 mol of any gas occupies a volume of 24.0 L at 20°C and 101 kPa.
What I have so far:
delta[H2]/(delta)t =
i.e. I have no idea what to do here...
1 answer