A mixture of 0.6 mole of al and 0.9 mole of cl2 are allowed to react as 2al+3cl2=2al2o3 . Calculate mass of reactant left unreacted. (al=27gmol-1 , cl=35.5gmol-1 )

You would do well to find the caps key and use it for the formulas. What you have written is complete nonsense but I can translate it.

Using the coefficients in the balanced equation, convert mols Al to mols Al2O3.
Do the same to convert mols Cl2 to mols Al2O3.
It is likely that these two values for mols Al2O3 will not agree; in limiting reagent problems the SMALLER value is always the correct one produced and the reagent producing that value is the limiting reagent (LR). Use the coefficients in the balanced equation to convert mols of the LR to mols of the OR(other reagent).
Subtract mols of OR initially - mols OR used to obtain moles OR remaining. Convert to grams by g = mols x molar mass. Post your work if you get stuck.