Stoechiométrie - 2 | Calculs chimiques

Equation: 2 Na (s) + Cl₂ (g) → 2 NaCl (s) 2 moles of Na form 2 moles of NaCl ⇒ n _{consumed Na} = n _{produced NaCl}
n _{consumed Na} = $\frac{{\mathrm{m}}_{\mathrm{Na}}}{{\mathrm{M}}_{\mathrm{Na}}}$ ⇒ n _{consumed Na} = $\frac{5.00}{23.0}$ = 2.17 x 10^-1 mol ⇒ n _{produced NaCl} = 2.17 x 10^-1 mol n _{produced NaCl} = n_NaCl = $\frac{{\mathrm{m}}_{\mathrm{NaCl}}}{{\mathrm{M}}_{\mathrm{NaCl}}}$ ⇒ m_NaCl = n _{produced NaCl} x M_NaCl ⇒ m_NaCl = 12.7 g
1 mole of Cl₂ is necessary to form 2 moles of NaCl ⇒ n_{required Cl2} = $\frac{{\mathrm{n}}_{\mathrm{produced} \mathrm{NaCl}}}{2}$ n_{required Cl2} = 1.09 x 10^-1 mol n_{required Cl2} = $\frac{{\mathrm{m}}_{\mathrm{Cl}2}}{{\mathrm{M}}_{\mathrm{Cl}2}}$ ⇒ m_Cl2 = n _{required Cl2} x M_Cl2 ⇒ m_Cl2 = 7.71 g