Question

How many moles of Al(Cn)3 are in 217g of the compound

Answer 1

Molar mass Al(CN)3 =>  105 g/mol

1 mole -------------  105 g
? mole ------------- 217 g

217 x 1 / 105 => 2.067 moles

Answer 2

Answer:

2.086 moles of Al(Cn)₃ are in 217g of the compound

Explanation:

To know the amount of moles of Al(CN)₃ in 217 g of the compound, you must know the molar mass of the compound. For that, you must know the atomic mass of each element that forms the compound and its relative abundance.

The atomic mass of each compound is obtained from the periodic table of the elements:

• Al: 27 g/mol
• C: 12 g/mol
• N: 14 g/mol

Considering relative abundance of each element in the compound, the molar mass is:

Al(CN)₃= 27 g/mol + 3*( 12 g/mol + 14 g/mol)= 105 g/mol

Now you apply a rule of three as follows: if 105 g are present in 1 mol of the compound, 217 g are present in how many moles of Al(CN)₃?

$moles=\frac{217 g*1 mole}{105 g}$

moles= 2.086

2.086 moles of Al(Cn)₃ are in 217g of the compound