If you have 35.0g of na and 100.0g of cl2, what is the maximum amount of nacl that you can produce? 2na + cl2 → 2nacl 57.7 g 165 g 75.8g 89.0 g

89.0 g.

Explanation:

2Na + Cl₂ → 2NaCl,

It is clear that 2 mol of Na react with 1 mol of Cl₂ to produce 2 mol of NaCl.

no. of moles of Na = mass/molar mass = (35.0 g)/(22.989 g/mol) = 1.522 mol.

no. of moles of Cl₂ = mass/molar mass = (100.0 g)/(70.906 g/mol) = 1.41 mol.

From the balanced equation: Na reacts with Cl₂ with (2: 1) molar ratio.

∴ 1.522 mol of Na "limiting reactant" reacts completely with 0.761 mol of Cl₂ "excess reactant".

2 mol of Na produces → 2 mol of NaCl, from stichiometry.

∴ 1.522 mol of Na produces → 1.522 mol of NaCl.

∴ The mass of the produced salt (NaCl) = (no. of moles)(molar mass) = (1.522 mol)(58.44 g/mol) = 88.95 g ≅ 89.0 g.

i'm assuming that the correct answer is indeed the first one/ "a"..

did anyone actually do this? like if anyone has submitted this i rly need .