Question

1)calculati masa azotului cu volumul 30 litri in conditii normale 2)ce volum ocupa clorura de hidrogen cu masa 14,6 grame in conditii normale?

Answer 1

1) p=1atm
T=273K
MN2=28g/mol
R=0,082

pV=m/MRT==>m=pVM/RT=1*30*28/0,082*273=840/22,386=37,523g N2

sau n=V/Vm=30/22,4=133928 moli N2

n=m/M==>m=n*M=1,33928*28=37,5g N2

2) MHCl=36,5g/mol

n=m/M=14,6/36,5=0,4 moli HCl

pV=nRT==>V=nRT/p=0,4*0,082*273/1=8,9544g HCl