Question

A student prepared some deep ruby crystals of ''chrome alum'' by treating a solution of potassium dichromate(VI) in dilute sulphuric acid with an excess of ethanol, and then crystallising the solution formed. What is the the maximum mass of chrome alum crystals that could be made from 2.94g of potassium dichromate(VI).

K2Cr2O7 + 4H2SO4 + 3CH3CH2OH --> K2SO4 + CR2(SO4)3 + 7H2O + 3CH3CHO

K2SO4 + CR2(SO4)3 + 24H2O --> K2SO4.CR2(SO403.24H2O

Answer 1

$$n = m/M$$

$$n(K_2Cr_2O_7) = {2.94g \over 294g/mol} = 0.01mol$$

$$n(K_2Cr_2O_7) = n(K_2SO_4) = n(K_2SO_4*Cr_2(SO_4)_3*24H_2O)$$

$$m = nM$$

$$m(K_2SO_4*Cr_2(SO_4)_3*24H_2O) = 0.01mol*998g/mol = 9.98g$$