Question

how many grams of CoCl2 are required to produce 5.30 x 10^23 molecules of chlorine

Answer 1

$$n = N/N_A = 5.30*10^{23}/6.02*10^{23} = 0.88 mol$$

$$n(CoCl_2) = n(Cl_2)$$

$$m(CoCl_2) = nM = 0.88*130 = 114.4 g$$