# Question #fc962

Jun 13, 2017

14,5% of Zn

#### Explanation:

From the gasses ideal Law and by consedering that 760 mmHg = 1 atm we have $n \left({H}_{2}\right) = \frac{P V}{R T} = \frac{1 a t m 0 , 328 L}{\left(0 , 082 \frac{L a t m}{m o l K}\right) 300 K} = 0 , 0133 m o l$.

These mol of ${H}_{2}$ comes from the reaction:
$Z n + {H}_{2} S {O}_{4} = Z n S {O}_{4} + {H}_{2}$
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.