Question

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

Answer 1

Molar mass of CaCl2.2H2O,

MM = 1*MM(Ca) + 2*MM(Cl) + 4*MM(H) + 2*MM(O)

= 1*40.08 + 2*35.45 + 4*1.008 + 2*16.0

= 147.012 g/mol

mass of CaCl2.2H2O = 1.5 g

mol of CaCl2.2H2O = (mass)/(molar mass)

= 1.5/1.47*10^2

= 1.02*10^-2 mol

According to balanced equation

mol of CaCl2 formed = moles of CaCl2.2H2O

= 1.02*10^-2 mol

Molar mass of CaCl2,

MM = 1*MM(Ca) + 2*MM(Cl)

= 1*40.08 + 2*35.45

= 110.98 g/mol

mass of CaCl2 = number of mol * molar mass

= 1.02*10^-2*1.11*10^2

= 1.132 g

Answer: 1.13 g