Question

a) If a piece of aluminum is placed in a solution of zince nitrate the result will be? why?

b)What mass of Ca is needed to produce 1.00g of CaCl? why?

c) Calculate the number of grams in 4.97 x 10^22 molecules of CO2? why?

Answer 1

A)

Al3+ + 3 e− ---> Al(s)    E0 = −1.662 V

Zn2+ + 2 e− ---> Zn(s)        E0 = −0.7618 v

as the reduction of potential of Al is lessthan Zn. Al gets oxidation and the Zn^2+(aq) gets reduction

gives formation Al-metal,

3Zn2+(aq) + 2Al(S) - --> 2Al^3+(aq) + 3Zn(s)

b) 1 mole CaCl2 = 1 mol Ca

NO of mol of CaCl2 = w/mwt = 1/111 = 0.009 mol

no of molo of Ca REQUIRED = 0.009 mol

mass of ca = 0.009*40 = 0.36 g

C) 1 mol CO2 = 6.023*10^23 molecules

   no of mol of CO2 = (4.97*10^22)/(6.023*10^23) = 0.0825 mol

   mass of CO2 = 0.0825*44 = 3.63 g