Question #fc962

1 Answer
Jun 13, 2017

14,5% of Zn


From the gasses ideal Law and by consedering that 760 mmHg = 1 atm we have #n(H_2) = (PV)/(RT)= (1 atm 0,328L)/ ((0,082 (L atm )/(mol K)) 300K)= 0,0133 mol#.

These mol of #H_2# comes from the reaction:
#Zn + H_2SO_4 = ZnSO_4 + H_2 #
from which you see that need 1 mol of Zn to form 1 mol of #H_2# so need 0,0133 mol of Zn to form 0,0133 mol of #H_2#
If 1 mol of atoms of Zn weigh 65,4 g, 0,0133 mol of Zn weigh 0,872 g that in 6 g of alloy are 14,5% of the brass.
That is a theoretical reaction because if Zn react quite quantitatively with sulfuric acid, Cu could reacts partially especially if sulfuric acid is warm and concentrated.