Question

$A{l}_2{O}_3+NaOH+H_{2}O\left(l\right)\to NaAl\left(OH{)}_4\right(aq)$

Moles of $NaOH$ required to balance equation are :

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (B)

$2$

${Al}_2{O}_3+NaOH+H_{2}O⟶NaAl(OH{)}_4$

Firstlyy balancing $Al$ on both sides.

${Al}_2{O}_3+NaOH+H_{2}O⟶2NaAl(OH{)}_4$

Now, balancing $Na$ on both sides.

${Al}_2{O}_3+2NaOH+H_{2}O⟶2NaAl(OH{)}_4$

Now balancing $H$ and $O$ on the both sides.

${Al}_2{O}_3+2NaOH+3H_{2}O⟶2NaAl(OH{)}_4$

$\therefore$ $Moles$ of $NaOH$ required to balance the equation $=2$ $Moles.$