225 mL of 0.445M H2SO4 solution is added 32.3 g Aluminium metals and reacts to completion. How many Moles of H2 are Produced ?

Question

225 mL of 0.445M H2SO4 solution is added 32.3 g Aluminium metals and reacts to completion. How many Moles of H2 are Produced ?

 

Al(s)+H2SO4(aq)----> Al2(SO4)3(aq)+H2(g)

Answer 1

1. Balance the reaction,

 

2Al(s)+3H2SO4(aq)----> Al2(SO4)3(aq)+3H2(g)

 

2. compare the amount (in moles) of both reactants,

 

Al  is   32.3g , which is slightly more than 1 mol of  Al.   

H2SO4  is 0.445 x 225/1000 mol,

 

3.  Evaluate mol to mol ratio of reactants,

 

Al:H2SO4 = 2 : 3  which implies H2SO4 is the reaction limiting, and is totally consumed upon reaction completion.

 

4.  Here, H2SO4 and H2 have the same proportion  in the balanced equation (factor =3),  you could figure out the rest.

Answer 2

Firs write out the chemical equation

 

Al(s)+H2SO4(aq)----> Al2(SO4)3(aq)+H2(g)

 

Then balance

 

2Al(s)+3H2SO4(aq)----> Al2(SO4)3(aq)+3H2(g)

 

Then do stoichiometry for both to find the limiting reactant and the theorectical yield (lowest number at the end)

 

starting with the singular unit number

 

32g Al x ( 1 mol Al/ 27 g Al) x (3 mol H2 / 2 mole Al) =  # mol H2

 

22mL x ( 1 L/ 1000mL) x ( 0.445 mol H2SO4/ 1L) x (3 mol H2/ 3 mol H2SO4) = # mol H2

 

The lowest number should be your theoretical yield and thus your answer.