Question

A solid piece of aluminum (51.0 g) was added to a solution of sodium hydroxide (84.1 g) in water, A balanced equation for this reaction is shown below:
2 NaOH(aq)+ 2 Al(s)+ 2 H2O → 2 NaAlO2(aq)+ 3 H2(g)

(a) Which reagent is completely consumed by the reaction?
(b) After the reaction is completed, what is the mass of the reagent that remains? (c) What mass of hydrogen gas is produced?

Answer 1

$M_{Al}=26,98\frac{g}{mol}\\ m=51g\\\\ n=\frac{m}{M_{Al}}=\frac{51g}{26,97\frac{g}{mol}}\approx1,89mol\\\\\\ M_{NaOH}=39,4\frac{g}{mol}\\ m=84,1g\\\\ n=\frac{m}{M_{NaOH}}=\frac{84,1g}{39,4\frac{g}{mol}}\approx2,14mol$

2 NaOH(aq)+ 2 Al(s)+ 2 H₂O → 2 NaAlO₂(aq)+ 3 H₂(g)
  2mol      :     2mol           :                                  3mol
2,14mol   :     1,89mol      :                                  2,835mol
remains         completely consumed
2,14-1,89=0,25mol


A) Al

B)  

$M_{NaOH}=39,4\frac{g}{mol}\\ n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g$

C)
$n=2,835mol\\ M_{H_{2}}=2,02\frac{g}{mol} \Rightarrow \ \ \ m=n*M_{H_{2}}=2,835mol*2,02\frac{g}{mol} \approx 5,73g$