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What is Chemical Reaction?
A chemical reaction is the chemical transformation of one set of chemical components into another.
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Mass of aluminium sulfide is 158g
Mass of water is 131g
The chemical reaction: [tex]Al_{2}S_{3} +H_{2}O _\to Al(OH)_{3} + H_{2}S[/tex]
First, balance the chemical equation
[tex]Al_{2}S_{3} + 6H_{2}O \to 2Al(OH)_{3} + 3H_{2}S[/tex]
Aluminium sulfide has a molar mass of 150.16 g/mol and water has a molar mass of 18.02 g/mol. As a result, the moles of aluminum sulfide are computed as follows:
[tex]n_{Al_{2}S_{3} } = \frac{Mass}{Molar mass}\\n_{Al_{2} S_{3} } = \frac{158g}{150.16g/mol} \\n_{Al_{2}S_{3} }=1.05 mol[/tex]
From the chemical reaction , the ratio of molar is 3mol [tex]H_{2}S/1 mol Al_{2}S_{3}.[/tex] So, the moles of hydrogen sulfide are:
[tex]n_{H_{2} O} =\frac{131g}{18.02g/mol}[/tex]
= 7.26mol
From the chemical reaction, the molar ratio is 3 mol [tex]H_{2}S/6[/tex] mol [tex]H_{2}O.[/tex] So, the moles of hydrogen sulfide are:
Moles of [tex]H_{2}S[/tex] formed = 7.26 mol [tex]H_{2}O \times \frac{3 mol H_{2}S }{6 mol H_{2} O} }[/tex]
Th liming reactant is[tex]Al_{2}S_{3}[/tex] beacuse the mass of [tex]Al_{2}S_{3}[/tex] forms less product than water. Therefore, the maximum number of moles of [tex]H_{2}S[/tex] is 3.15 mol. We know that molar mass of [tex]H_{2}S[/tex] is 34.10g/mol. So, the maximum mass of [tex]H_{2}S[/tex] formed is,
[tex]m_{H_{2}S } = n_{H_{2}S } \times[/tex] Molar mass of [tex]H_{2}S[/tex]
= 3.15 mol [tex]\times[/tex] 34.10g/mol
= 107.4g
Now, multiplying the number of moles of [tex]Al_{2}S_{3}[/tex] by the molar ratio between [tex]Al_2S_3[/tex] and [tex]H_2O[/tex] which is 6mol [tex]H_2O/1mol[/tex] [tex]Al_2S_3[/tex] we get the number of moles of [tex]H_2O[/tex] reacted.
Moles of [tex]H_2O[/tex] reacted = 1.05 mol [tex]Al_{2}S_3 \times \frac{6 mol H_2O}{1 mol Al_2S_3}[/tex]
= 6.31 mol [tex]H_2O[/tex]
The mass of [tex]H_2O[/tex] is,
[tex]m_{H_{2} O}[/tex] = 6.31 mol [tex]\times[/tex] 18.02g/ mol
= 114g
On subtracting, the mass of [tex]H_2O[/tex] reacted from the given mass of [tex]H_2O[/tex] is,
[tex]m_{H_2O}[/tex] = (131-114)g
= 17g
Hence, the excess remaining reactant is 17g
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