Question

You are given a 215 mg sample of a barium chloride hydrate BaCl2·6H2O and told to heat it in order to drive off all the waters of hydration. What mass of anhydrous salt will you be left with after cooling?

Answer 1

Molar mass of BaCl2.6H2O,

MM = 1*MM(Ba) + 2*MM(Cl) + 12*MM(H) + 6*MM(O)

= 1*137.3 + 2*35.45 + 12*1.008 + 6*16.0

= 316.296 g/mol

mass of BaCl2.6H2O = 0.215 g

mol of BaCl2.6H2O = (mass)/(molar mass)

= 0.215/3.163*10^2

= 6.797*10^-4 moL

According to balanced equation

mol of BaCl2 formed = moles of BaCl2(H2O)6

= 6.797*10^-4 mol

Molar mass of BaCl2,

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

= 1*137.3 + 2*35.45

= 208.2 g/mol

mass of BaCl2 = number of mol * molar mass

= 6.797*10^-4*2.082*10^2

= 0.142 g

= 142 mg

Answer: 142 mg