How much excess reactant remains from the reaction of 8.3 g of sodium and 4.5 g of chlorine? 2Na(s) + Cl2(9) —>2NaCl
A. 5.4 g Na
B. 8.2 g Cl2
C. 8.2 g Na
D. 5.4 g Cl2

Sodium:
moles = mass/RFM = 8.3/23 = 0.3608..

Chlorine:
moles = mass/RFM = 4.5/71 = 0.06

Ratio of Na to Cl is 2:1
So 0.06 x 2 = 0.12

So we know that the Na is in excess

Now we have to figure out by how much:

Mass = moles x RFM = 0.12 x RFM of 2NaCl = 0.12 x 58.5 = 7.4 (roughly)

Now to figure out by how much do 8.3 + 4.5 = 12.7 and do 12.7 - 7.4 = 5.3 (roughly)

So your answer is A

I’ve taken this before! the answer isa ^.^