For the reaction, calculate how many grams of the product form when 3.4 g of Cl2completely reacts.?

Molar mass of Cl₂ = 35.5×2 g/mol = 71.0 g/mol
Molar mass of NaCl = (23.0 + 35.5) g/mol = 58.5 g/mol

2Na(s) + Cl₂(g) → 2NaCl(s)
Mole ratio Cl₂ : NaCl = 1 : 2

No. of moles of Cl₂ reacted = (3.4 g) / (71.0 g/mol) = 0.0479 mol
No. of moles of NaCl formed = (0.0479 mol) × 2 = 0.0958 mol
Mass of NaCl formed = (0.0958 mol) × (58.5 g/mol) = 5.6 g

5.6g of NaCL if considering significant figures.
Here's my work

2Na(s) + Cl₂(g) → 2NaCl(s).

The molar mass of Cl₂ is 70.9 g/mol, so we have (3.4 g)/(70.9 g/mol) = 0.048 mol.
The mole ratio of Cl₂ to NaCl is 1:2, so 2(0.048) = 0.096 mol of NaCl is produced.
The molar mass of NaCl is 58.44 g/mol, so we have (0.096 mol)(58.44 g/mol) = 5.6 g.