How many moles are in 98.3 grams of aluminum hydroxide, al(oh)3?

We have to calculate the molar mass of  AL(OH)₃
Atomic mass (Al)=27 amu
Atomic mass (O)=16 amu
Atomic mass (H)=1 amu

molecular weight= 27 amu+3(16 amu + 1 amu) =78 amu.

Therefore, the molar mass of Al (OH)₃ is 78 g/ mol

Now, we calculate the number of moles in 98.3 g of aluminum hydroxide.

78 g1 mol
98.3 g  x

x=(98.3 g * 1 mol) / 78 g=1.26 moles.

1.26 moles.

It is called a neutralization reaction