Question

calculate how many moles of CaCl2•2H2O are present in 1.50 g of CaCl2•2H2O and then calculate how many moles of pure CaCl2 are present in the 1.50 g of CaCl2•2H2O.

Answer 1

`0.0102~mol~CaCl_2*2H_2O`

`0.0102~mol~CaCl_2`

Step-by-step explanation:

For this question, we have to start with the molar mass calculation of
`CaCl_2*2H_2O`. For this, we have to know the atomic mass of each atom:

O: 16 g/mol

Cl: 35.45 g/mol

H: 1 g/mol

Ca: 40 g/mol

If we take into account the amount of each atom in the formula we will have:

`(40*1)+(35.45*2)+(1*4)+(16*2)=~147.01~g/mol`

So, in 1 mol of
`CaCl_2*2H_2O` we will have 147.01 g. Now we can do the conversion:

`1.50~g~CaCl_2*2H_2O(1~mol~CaCl_2*2H_2O)/(147.01~g~CaCl_2*2H_2O)=0.0102~mol~CaCl_2*2H_2O`

Additionally, in 1 mol of
`CaCl_2*2H_2O` we will have 1 mol of
`CaCl_2`. Therefore, we have a 1:1 mol ratio . With this in mind, we will have the same number of moles for
`CaCl_2`

`0.0102~mol~CaCl_2*2H_2O=0.0102~mol~CaCl_2`

I hope it helps!