Calculate concentration

$c = \frac{n}{V}$

$m = n*M$

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Concentration of water in water

$1L = \frac{1000g}{18g/mol}=55$ mol

$\rho_{H_2O} = \frac{1000g}{L}=$

$C_{H_2O}=\frac{n}{V}=\frac{55}{1}=55$ mol/liter

Limiting factor

Smallest amount determines limiting reactant

Limitic reactant

Formation of NaCl

70 g $Na$ (s) = $n = \frac{m}{M} = \frac{70}{23} ≈ 3$ mol

70 g $Cl_2$(g) = $n = \frac{m}{M} = \frac{70}{71} ≈ 1$ → $2$ fordi 2

$2 Na + Cl_2 → 2 NaCl$

Electrolytes