rody2011
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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)

AWhich reagent is completely consumed by the reaction?
BAfter the reaction is completed, what is the mass of the reagent that remains? (c) What mass of hydrogen gas is produced?

Verified Answer :

Yipes

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

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

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