Under the action of hydrochloric acid on 5.5 g of a mixture of iron and aluminum, 4.48 liters of hydrogen were released

Under the action of hydrochloric acid on 5.5 g of a mixture of iron and aluminum, 4.48 liters of hydrogen were released. Determine the mass composition of the mixture.

Solution:

x y
Fe + 2 * H Cl = Fe Cl2 + H2
56 g 22.4 l

5 – x 4, 48 – y
2 * Al + 6 * H Cl = 2 * Al Cl3 + 3 * H2
2 * 27 g 67.2 l

then we get the system
x / 56 = y / 22.4
(5 – x) / 5 = (4, 48 – y) / 67, 2
hence x = 56 y / 22.4 = 2.5 y
we substitute it into the second equation, we get
(5 – 2.5 y) / 54 = (4.48 – y) = 67, 2
67, 2 * (5 – 2, 5 y) = 54 * (4, 48 – y)
67, 2 (2 – y) = 21, 6 * 4, 48 2 1, 6 * y
67, 2 * 2 – 21, 6 * 4, 4 8 = 45, 6 y
45.6 y = 37.632
y = 0.825
then x = 2.5 * 0.825 = 2.0 6 25 is the mass of iron
mass of aluminum 5 – x = 5 – 2, 0625 = 2, 9375
the amount of iron substance = 41, 25 percent
amount of aluminum substance = 58.7 5 percent