How many grams of Zn reacted IN Zinc Iodide reaction?

Molar mass of Zn = 65.38 g/mol
Molar mass of I₂ = 126.9 × 2 g/mol = 253.8 g/mol

Initial number of moles of Zn = (2.036 g) / (65.38 g/mol) = 0.03114 mol
Initial number of moles of I₂ = (1.972 g) / (253.8 g/mol) = 0.007770 mol

Balanced equation for the reaction:
Zn + I₂ → ZnI₂
Mole ratio Zn : I₂ = 1 : 1

If 0.007770 mol I₂ completely reacts, no. of moles of Zn needed = 0.007770 mol < 0.03114 mol
Zn is in excess, and thus I₂ is the limiting reactant/reagent.

Mass of Zn reacted = (2.036 - 1.536) g = 0.500 g
No. of moles of Zn reacted = (0.500 g) / (65.38 g/mol) = 0.007648 mol
No. of moles of I₂ reacted = 0.007648 mol
Mass of I₂ reacted = (0.007648 mol) × (253.8 g/mol) = 1.941 g

How many grams....

Zn(s) + I2(g) --> ZnI2(s)
2.036g 1.972g......?g

One way to solve a limiting reactant problem is to compute the amount of product using both reactants. The theoretical yield will be the lesser of the two, and that will tell you the limiting reactant as well.

2.036g Zn x (1 mol Zn / 65.38g Zn) x (1 mol ZnI2 / 1 mol Zn) x (322.18g ZnI2 / 1 mol ZnI2) = 10.03g ZnI2
1.972g I2 x (1 mol I2 / 253.8g I2) x (1 mol ZnI2 / 1 mol I2) x (322.18g ZnI2 / 1 mol ZnI2) = 2.503g ZnI2

The theoretical yield is 2.503g and I2 is the limiting reactant.

Compute theoretical amount of zinc which reacted:
1.972g I2 x (1 mol I2 / 253.8g I2) x (1 mol Zn / 1 mol I2) x (65.38g Zn / 1 mol Zn) = 0.5080g Zn reacted

Excess of zinc = 2.036g - 0.5080g = 1.528g (theoretical)

Now we have a grip on the reaction. But instead, you got an excess of 1.536g Zn.

=========== =========== =====
2.036g - 1.536g = 0.500g Zn reacted
=========== =========== =====

Actual yield
0.500g Zn x (1 mol Zn / 65.38g Zn) x (1 mol ZnI2 / 1 mol Zn) x (322.18g ZnI2 / 1 mol ZnI2) = 2.464g ZnI2 formed.