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

Answer:

$0.0102~mol~CaCl_2*2H_2O$

$0.0102~mol~CaCl_2$

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:

$1.50~g~CaCl_2*2H_2O\frac{1~mol~CaCl_2*2H_2O}{147.01~g~CaCl_2*2H_2O}=0.0102~mol~CaCl_2*2H_2O$

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!