In the reaction $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3$ , how many moles of $A l {(O H)}_{3}$ $Al(OH)_3$ can be made with 2.3 moles of $N a O H$ $NaOH$ and excess $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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In the reaction $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3$ , how many moles of $A l {(O H)}_{3}$ $Al(OH)_3$ can be made with 2.3 moles of $N a O H$ $NaOH$ and excess $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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Answer 1 · 2016-02-09T18:21:25

$0.767 m o l$ $0.767mol$

Explanation:

Since aluminium sulphate is in excess, it implies that sodium hydroxide is the limiting reactant and decides how much product is formed as it gets used up first.

The balanced chemical equation represents the mole ratio in which the chemicals react.

Since $6 m o l N a O H$ $6 mol NaOH$ produces $2 m o l A l {(O H)}_{3}$ $2molAl(OH)_3$ , ie 3 times less, it implies that $2, 3$ $2,3$ moles of $N a O H$ $NaOH$ will produce $\frac{2.3}{3} = 0.767$ $2.3/3=0.767$ moles of $A l {(O H)}_{3}$ $Al(OH)_3$ .

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In the reaction $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} {(S O_{4})}_{3}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3$ , how many moles of $A l {(O H)}_{3}$ $Al(OH)_3$ can be made with 2.3 moles of $N a O H$ $NaOH$ and excess $A l_{2} (S O_{4}) 3$ $Al_2(SO_4)3$ ?

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In the reaction Al2(SO4)3+6NaOH→2Al(OH)3+3Na2(SO4)3Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2(SO_4)_3, how many moles of Al(OH)3Al(OH)_3 can be made with 2.3 moles of NaOHNaOH and excess Al2(SO4)3Al_2(SO_4)3?

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