Answer
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Hint: We know that chemical equations in which the number of atoms of all the involved species is equal on the reactant and the product side are called balanced chemical equations.
Step by step answer:
We use chemical equations to represent chemical reactions by using the chemical formulae of all the involved species.
The aluminium oxide is specifically identified as aluminium (III) oxide with chemical formula ${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}$and sodium oxide based on the formula NaOH. When these two react, sodium aluminate is formed along with water.
So the general reaction of this as shown
below.${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}} + {\rm{NaOH}} \to {\rm{NaAl}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$
So, we have to balance the moles of individual elements present in both the sides, the reactant and the product sides for balancing the above reaction. This reaction can generally be balanced by one of the well-known methods that are hit and trial method.
So, in the reactant side's reaction, we have two aluminium atoms, four oxygen atoms, one sodium atom, and one hydrogen atom. Similarly, on the product side, we have one aluminium atom, three oxygen atoms, one sodium atom, and two hydrogen atoms.
Thus, with the help of the hit and trial method, we can say that there must be one mole of aluminium oxide that reacts with two moles of sodium hydroxide to form two moles of sodium aluminate and one mole of water.
${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}} + {\rm{2NaOH}} \to 2{\rm{NaAl}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$
Hence, the correct option is D.
Note: Thus, aluminium oxide is amphoteric, which means it acts both as acid and base depending upon the situation. Sodium hydroxide is a base.
Step by step answer:
We use chemical equations to represent chemical reactions by using the chemical formulae of all the involved species.
The aluminium oxide is specifically identified as aluminium (III) oxide with chemical formula ${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}}$and sodium oxide based on the formula NaOH. When these two react, sodium aluminate is formed along with water.
So the general reaction of this as shown
below.${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}} + {\rm{NaOH}} \to {\rm{NaAl}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$
So, we have to balance the moles of individual elements present in both the sides, the reactant and the product sides for balancing the above reaction. This reaction can generally be balanced by one of the well-known methods that are hit and trial method.
So, in the reactant side's reaction, we have two aluminium atoms, four oxygen atoms, one sodium atom, and one hydrogen atom. Similarly, on the product side, we have one aluminium atom, three oxygen atoms, one sodium atom, and two hydrogen atoms.
Thus, with the help of the hit and trial method, we can say that there must be one mole of aluminium oxide that reacts with two moles of sodium hydroxide to form two moles of sodium aluminate and one mole of water.
${\rm{A}}{{\rm{l}}_{\rm{2}}}{{\rm{O}}_{\rm{3}}} + {\rm{2NaOH}} \to 2{\rm{NaAl}}{{\rm{O}}_{\rm{2}}} + {{\rm{H}}_{\rm{2}}}{\rm{O}}$
Hence, the correct option is D.
Note: Thus, aluminium oxide is amphoteric, which means it acts both as acid and base depending upon the situation. Sodium hydroxide is a base.
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