Question

Consider the reaction below
Al2S3(s) + H2O(l) → Al(OH)3(s) + H2S(g)
If 15.0g of aluminum sulfide and 10.0g of water are allowed to react as above, and assuming a complete reaction
a. by calculation, find out which is the limiting reagent.
b. calculate the maximum mass of H2S which can be formed from these reagents.
c. calculate the mass of excess reagent remaining after the reaction is complete.
HELP PLEASE

Answer 1

hint:
the numbers of moles of the chemical agents involved, are:
\[\begin{gathered}
moles\;of\;A{l_2}{S_3} = \frac{{15}}{{150}} = 0.1 \hfill \\
\hfill \\
moles\;of\;{H_2}O = \frac{{10}}{{18}} = 0.56 \hfill \\
\end{gathered} \]

Answer 2

what about B tho