Answer:
m H2(g) = 2.241 g H2(g)
Explanation:
- 2Al(s) + 3H2SO4(aq) → Al2(SO4)3(aq) + 3 H2(g)
limit reagent:
∴ Mw Al = 26.982 g/mol
∴ Mw H2SO4 = 98.0785 g/mol
⇒ n Al = (20 g Al)×(mol/26.982 g) = 0.7412 mol Al
⇒ n H2SO4 = ( 115 g H2SO4 )×(mol/98.0785 g) = 1.173 mol H2SO4
⇒ n H2 = (0.7412 mol Al)×(3 mol H2/ 2 mol Al) = 1.112 mol H2
∴ Mw H2 = 2.016 g/mol
⇒ g H2 = (1.112 mol H2)×(2.016 g/mol) = 2.241 g H2
The final step in a typical titration, that is here an acid base one would be to finally find the concentration of your unknown substance whether that be the acid or the base. The other steps are used before this to come to the correct calculation and conclusion.
The notion <span>an empty balloon have precisely the same apparent weight on a scale as a balloon filled with air depends on the diameter of the balloon. The weight of the balloon filled with air is equal to the mass of the balloon and the mass of the air inside. The mass of air inside is equal to the density of air multiplied by the volume of the balloon. If the balloon is large, then the two masses are equal whereas if not, the mass of air inside the inflation is neglible</span>
