In the equation $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} S O_{4}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4$ , what is the mole ratio of $N a O H$ $NaOH$ to $A l {(O H)}_{3}$ $Al(OH)_3$ ?

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In the equation $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} S O_{4}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4$ , what is the mole ratio of $N a O H$ $NaOH$ to $A l {(O H)}_{3}$ $Al(OH)_3$ ?

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Answer 1 · 2016-05-28T14:56:08

3:1

Explanation:

The balanced equation will tell the stoichiometric ratio of reactants and products. Here in the balanced equation the coefficient before $N a O H$ $NaOH$ is 6 and that before $A l {(O H)}_{3}$ $Al(OH)_3$ is 2.
So the required mole ratio is 6:2=3:1

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In the equation $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} S O_{4}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4$ , what is the mole ratio of $N a O H$ $NaOH$ to $A l {(O H)}_{3}$ $Al(OH)_3$ ?

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In the equation Al2(SO4)3+6NaOH→2Al(OH)3+3Na2SO4Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4, what is the mole ratio of NaOHNaOH to Al(OH)3Al(OH)_3?

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In the equation $A l_{2} {(S O_{4})}_{3} + 6 N a O H \to 2 A l {(O H)}_{3} + 3 N a_{2} S O_{4}$ $Al_2(SO_4)_3 + 6NaOH -> 2Al(OH)_3 + 3Na_2SO_4$ , what is the mole ratio of $N a O H$ $NaOH$ to $A l {(O H)}_{3}$ $Al(OH)_3$ ?