Calculate the maximum numbers of moles and grams of H2S

Chapter 3, Problem 51P

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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?

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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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