To find the limiting and the excess reactant we will need first to set up the balanced reaction:
Al + 3NaCl -> AlCl3 + 3Na
Now with the balanced reaction, we will also need the molar mass for Al and NaCl, which are:
Al = 26.98 g/mol
NaCl = 58.44 g/mol
Then we will need to find how many moles do 10 grams represents in each case, and how much of the other reactant it would be needed to react:
26.98 g = 1 mol
10 g = x moles
x = 0.37 moles
According to the molar ratio in the reaction, we will always have 1:3 ratio, which means that in order to react 1 mol of Al, we will need 3 moles of NaCl, using that in an equation:
1 Al = 3 NaCl
0.37 = x NaCl
x = 1.11 moles of NaCl
But how many grams would 1.11 moles of NaCl represent? We know is more than 58.44 grams, since in order to have that only 1 mol would be needed, so now we know that the limiting reactant is the NaCl and we have an excess of Al, but let's see how much excess:
58.44 g = 1 mol
10 g = x moles
x = 0.17 moles of NaCl, which means 0.056 moles of Al, since we are dividing 0.17 by 3
26. 98 g = 1 mol
x grams = 0.056 moles
x = 1.51 grams
Therefore, limiting reactant = NaCl, excess reactant = Al, left over = 10g - 1.51g = 8.49 grams