Answer to Question #140200 in General Chemistry for yara

Question

How many moles are in 98.3 g of aluminum hydroxide, Al(OH)3?

Answer 1

Q140200

How many moles are in 98.3 g of aluminum hydroxide, Al(OH)₃ ?

Solution :

Step 1 : Find the number of atoms of each element in Al(OH)₃ .  

There is no subscript given to Al. This means there is 1 atom of Al.

A subscript 3 is written outside the parenthesis, So multiply the atoms inside the parenthesis by 3.

Al = 1, O = 3 and H = 3.

Step 2: Find the molar mass of Al(OH)₃

Atomic mass of Al = 26.982 g/mol, Atomic mass of O = 15.999 g/mol,

Atomic mass of H = 1.00794 g/mol

Molar mass of Al(OH)₃ = 1 * atomic mass of Al + 3 * atomic mass of O + 3 * atomic mass of H

= 1 * 26.98g/mol + 3 * 15.999 g/mol + 3 * 1.00794 g/mol

= 26.982 g/mol + 47.997 g/mol + 3.0238 g/mol

= 78.0027 g/mol

Step 3 : Convert 98.3g of Al(OH)₃ to moles using its molar mass.

Moles of Al(OH)₃ = 98.3g of Al(OH)₃ * 1 mol Al(OH)₃ /78.0027 g Al(OH)₃

= 1.260 mol of Al(OH)₃ .

In the question we are given 98.3g in 3 significant figures, So our final answer must also be in 3 significant figures.

Hence there is 1.26 mol of Al(OH)₃ in the given mass of Al(OH)₃ .