[genie]

Could anybody help me answer these questions, please ?
Given the reaction

Mg + 2HCl ----> MgCl2 + H2

1/If an undetected bubble of air were trapped inside the gas collection tube, would the your value of R be erroneously high,  erroneously low, or be unaffected by the bubble of air ? Explain

2/ If you hadn't added quite enough HCl to consume all of the magnesium metal, would your value of R be erroneously high,  erroneously low, or be unaffected by not adding enough HCl ? Explain

P/S : R is the value of the ideal gas law constant  and to be 0.08206L.atm/Kmol
and PV = nRT

Thank you so much

[genie]

Thank Borek for reminding me the forum rules

I had solved these problems

We know PV= nRT
then
1/ If an undetected bubble of air were trapped inside the gas collection tube, that means the volume V is increased, and then R is also increased => R would be erroneously high

2/If you hadn't added quite enough HCl to consume all of the magnesium metal, that means the moles of reaction was smaller than the moles of be given, and then R is decreased => R would be erroneously low

Does that work ?
Gen

[Borek]

Looks OK to me.

Mathematically - if you look at PV=nRT - you see that it can be written as

R=kV

or

R=k/n

(where k is some constant, if we can guarantee everything else to be not changing during experiment, we can combine all those different variables together into one number)

Now it is obvious that calculated R is directly proportional to measured V - so too high V means too high R, and that calculated R is inveresely proportional to measured n - that is, lower n, higher R value. That's what you have stated.