When aluminum reacts with sulfuric acid, aluminum sulfate and hydrogen are formed. Calculate:

When aluminum reacts with sulfuric acid, aluminum sulfate and hydrogen are formed. Calculate: a) the volume of hydrogen, if the mass of aluminum is 5.4 g; b) the mass of aluminum sulfate, if the mass of aluminum with impurities is 20 g, with a mass fraction of impurities of 4%

а)
Given:
m (Al) = 5.4g
Calculate:
V (H2) -?
2 mol 3 mol
2Al + 3H2SO4  Al2 (SO4) 3 + 3H2
___
0.2 mol? mole

M (Al) = 27g / mol
We find the amount of the substance of Aluminum:
v (Al) = 5.4g / 27g / mol = 0.2 mol
Based on the reaction equation, it can be seen that the amount of aluminum is related to the amount of hydrogen as 2: 3. Hence :
v (H2) = 0.3 mol
MOLAR VOLUME OF GASES under normal conditions = 22.4 l / mol =>
V (H2) = 0.3 mol * 22.4 l / mol = 6.72 l

b)
Given:
m (Al) = 20g
w (impurities) = 4%
Calculate:
m (Al2 (SO4) 3) -?

2 mol 1 mol
2Al + 3H2SO4  Al2 (SO4) 3 + 3H2
___
0.2 mol? mole

w (Al) = 100% -4% = 96%
m (Al) = 20g * 0.96 = 19.2g
M (Al) = 27g / mol
v (Al) = 19.2g / 27g / mol = 0.7mol
Based on the reaction equation, it can be seen that the amount of aluminum substance refers to the amount of aluminum sulfate substance as 2: 1. I.e
v (Al2 (SO4) 3) -? – 1/2 v (Al) = 0.7 mol / 2 = 0.35 mol
M (Al2 (SO4) 3) = 27 * 2 + (32 + 16 * 4) * 3 = 342g / mol
m (Al2 (SO4) 3) = 0.35mol * 342g / mol = 119.7g