Calculate the maximum numbers of moles and grams of H2S that can form when 158 g of aluminum sulfide reacts with 131 g of water:

$$

\mathrm{Al}_{2} \mathrm{S}_{3}+\mathrm{H}_{2} \mathrm{O} \longrightarrow \mathrm{Al}(\mathrm{OH})_{3}+\mathrm{H}_{2} \mathrm{S}[\text { unbalanced }]

$$

How many grams of the excess reactant remain?

$\mathrm{n}\left(\mathrm{H}_{2} \mathrm{S}\right)=3^{*} \mathrm{n}\left(\mathrm{Al}_{2} \mathrm{A}_{3}\right)=3,15 \mathrm{mol}$

$\mathrm{m}\left(\mathrm{H}_{2} \mathrm{S}\right)=107,29 \mathrm{g}$

$\mathrm{m}\left(\mathrm{H}_{2} \mathrm{O},\right.$ excess $)=18 \mathrm{g} / \mathrm{mol}^{*} 0,98 \mathrm{mol}=17,64 \mathrm{g}$

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