In order to answer this question we will need a few steps
First one will be to set up the properly balanced equation:
2 Na + Cl2 -> 2 NaCl
20 grams of Na
10 grams of Cl2
From this balanced equation we see that the molar ratio between Na and Cl2 is 2:1, that means that for every 2 moles of Na, we will need 1 mol of Cl2 in order to proceed with the reaction
Now we need to identify the limiting and excess reactant, we will do that by checking how many moles of each element we have and how much we should have in order to properly react with the other element, let's check it:
Molar mass for Na is 23g/mol
23g = 1 mol
20g = x moles
x = 0.89 moles of Na in 20 grams
So if we have 0.89 moles of Na, we should have 0.44 moles of Cl2, but we don't know if we have this amount of moles of Cl2 in 10 grams of it, let's check
Molar mass for Cl2 is 71g/mol
71g = 1 mol
10g = x moles
x = 0.14 moles
We have fewer moles than we actually needed, let's do the same step but now focusing on Cl2
Cl2 in 10 grams has 0.14 moles, therefore we will have 0.28 moles of Na, remember the molar ratio 2:1
23g = 1 mol
x grams = 0.28 moles
x = 6.44 grams of Na
If we have 10 grams of Cl2, we would only need 6.44 grams of Na, since we have more available, we can say that Na is in excess and Cl2 is the limiting reactant
To find out the mass of NaCl produced from 10 grams of Cl2 (we have to use the limiting reactant), NaCl has a molar mass of 58.44g/mol and the molar ratio will be again 2:1, one mol of Cl2 and 2 moles of NaCl, if we have 0.14 moles of Cl2, therefore we will have 0.28 moles of NaCl
58.44 g = 1 mol
x grams = 0.28 moles
x = 16.4 grams, the value in the options is a little higher, but not too much higher, that must be due some rounding up
Letter A