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ch4aika [34]
2 years ago
9

Will give brainiest

Chemistry
1 answer:
zimovet [89]2 years ago
3 0
MM=97.10 that’s the answer
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M_{Al}=26,98\frac{g}{mol}\\
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<span>2 NaOH(aq)+ 2 Al(s)+ 2 H</span>₂<span>O → 2 NaAlO</span>₂<span>(aq)+ 3 H</span>₂<span>(g)
</span>  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}\\&#10;n=0,25mol \Rightarrow \ \ \ m=n*M_{NaOH}=0,25mol*39,4\frac{g}{mol}=9,85g

C)
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3 years ago
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