if you want to make 2 L of Hydrogen gas at STP, what mass of Mg you want to react with excess amount of Hcl?

Mg+2Hcl----> MgCl2+H2
Limiting Reactant-- Mg
Molar mass of Mg:  24.305
1 mole of Mg produce 1 mole of H2 as per above equation.
PV=nRT
PV=(m/M)RT
mass of Mg is the question. So:

m=PVM/RT
At STP========>P=1 atm       T=298 K    R= 0.0821 atm?L/mol?K
m=1(2L)(24.305)/(.0821)(298)=1.987 gms of Mg

Is this correct? Please advise.

Quote from: prasamb@yahoo.com on November 30, 2022, 09:30:59 AM

PV=nRT
PV=(m/M)RT

m=PVM/RT
At STP========>P=1 atm       T=298 K    R= 0.0821 atm?L/mol?K
m=1(2L)(24.305)/(.0821)(298)=1.987 gms of Mg

Is this correct? Please advise.

Why are using ideal gas law on solid Mg?
It is applicable only on gases.STP condition is only for Hydrogen gas.
At STP all gases have 22.4L for 1 mole of gas.
It means you can convert
2L of H2 gas ----> moles of H2 -----> moles of Mg ------> grams of Mg .

2L H2 x 1mole H2/ 22.4L of H2    x 1mole of Mg / 1 mole of H2   x 24.30g Mg / 1 mole Mg

Thanks Uma

Hello Uma: Have an additional question in this problem.
if you want to make 2 L of Hydrogen gas at STP, what mass of Mg you want to react with excess amount of Hcl

Does this clearly say, HCL is on excess and we are supposed to use Mg as Limiting Reactant for our calculation. Kind of a Hint.
Am I correct?

Thanks
Prasanna

Yes it means Mg is the limiting reactant.
But if H₂ is already given as 2L at STP then it means you know amount of H₂ and they want you to calculate Mg..