Write the reaction of aluminum with caustic s | vedclass

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Question

(C) The balanced chemical equation for the reaction of aluminum with caustic soda $(NaOH)$ is:$2Al(s) + 2NaOH(aq) + 6H_2O(l) \rightarrow 2Na[Al(OH)_4]

Vedclass · Accepted Answer

$3$

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(C) The balanced chemical equation for the reaction of aluminum with caustic soda $(NaOH)$ is:$2Al(s) + 2NaOH(aq) + 6H_2O(l) ightarrow 2Na[Al(OH)_4](aq) + 3H_2(g)$Step $1$: Calculate the moles of $Al$:$Moles \ of \ Al = \frac{	ext{Given mass}}{	ext{Molar mass}} = \frac{54 \ g}{27 \ g/mol} = 2 \ mol$Step $2$: Use stoichiometry:From the balanced equation,$2 \ mol$ of $Al$ produces $3 \ mol$ of $H_2$.Therefore,$54 \ g$ $(2 \ mol)$ of $Al$ will produce $3 \ mol$ of $H_2$ gas at $STP$.

Write the reaction of aluminum with caustic soda and determine how many moles of $H_2$ gas will be produced at $STP$ from $54 \ g$ of $Al$. (in $mol$)

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