Question

3NaOH + Al(NO3)3 > 3NaNO3+ Al(OH)3, What volume of 0.300M NaOH is required to react completely with 2.50 g Al(NO3)3?

Answer 1

Moles of $Al(NO_3)_3= \frac{2.50}{213}=0.012moles$

Moles of $NaOH=3×0.012=0.008moles$

Volume $=\frac{0.008}{0.300}=0.027 litres$