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Calculate the maximum numbers of moles and grams of H2S
Chapter 3, Problem 51P(choose chapter or problem)
Calculate the maximum numbers of moles and grams of \(\mathrm{H}_{2} \mathrm{S}\) that can form when \(158 g\) of aluminum sulfide reacts with \(131 g\) of water:
\(\mathrm{A}1_2\mathrm{S}_3+\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{A}1(\mathrm{OH})_3+\mathrm{H}_2\mathrm{S}\) [unbalanced]
What mass of the excess reactant remains?
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QUESTION:
Calculate the maximum numbers of moles and grams of \(\mathrm{H}_{2} \mathrm{S}\) that can form when \(158 g\) of aluminum sulfide reacts with \(131 g\) of water:
\(\mathrm{A}1_2\mathrm{S}_3+\mathrm{H}_2\mathrm{O}\rightarrow\mathrm{A}1(\mathrm{OH})_3+\mathrm{H}_2\mathrm{S}\) [unbalanced]
What mass of the excess reactant remains?
ANSWER:
Step 1 of 6
Here, we are going to calculate the given quantities of \(\mathrm{H}_2\mathrm{S}\).
The balanced chemical equation is given below:
\(\mathrm{Al}_2\mathrm{S}_3+6\mathrm{H}_2\mathrm{O}\rightarrow2\mathrm{A}1(\mathrm{OH})_3+3\mathrm{H}_2\mathrm{S}\)
If we consider \(\mathrm{H}_2\mathrm{O}\) is the limiting agent, then the moles of \(\mathrm{H}_2\mathrm{S}\) from the \(\mathrm{Al}_2\mathrm{S}_3\) will be
Given that,
The mass of \(\mathrm{Al}_2\mathrm{S}_3=158\ \mathrm{g}\)
The molar mass of \(\mathrm{Al}_2\mathrm{S}_3=150.17\ \mathrm{g/mol}\)
Thus,
The moles of \(\mathrm{Al}_2\mathrm{S}_3=\left(158\mathrm{\ g}\mathrm{\ Al}_2\mathrm{S}_3\right)\times\frac{1.0\mathrm{\ mol}\text{ of }\mathrm{Al}_2\mathrm{S}_3}{150.17\mathrm{\ g}\mathrm{\ Al}_2\mathrm{S}_3}=1.052\mathrm{\ mol}\mathrm{\ Al}_2\mathrm{S}_3\)
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