Balance this equation- NaCl ----> Na + Cl Then answer (with work if possible) how many grams of Na we would have if we started with 139 g of NaCl. I so far get to here: 2NaCl ----> 2Na + Cl_2 (139gNaCl/1)(1MoleNaCl/58.44G)(22.99gNa/1Mole)?

2NaCl ---> 2Na^(+) + 2Cl^(-)

------This is correct, save ionization.

139 g NaCl (1 mole NaCl/58.44 g)(2 mole Na/2 mole NaCl)(22.99 g/1 mole Na)

=

------Looks right to me, except you left out the molar ratio step, though it is superfluous ( one to one ) it should be included to be formal. You have plugged, now chug!!