= Solution
The reaction consumes one <mole> of CF for each <mole> of <electrons>. Using $M_{\rm CF}=12.01+19.00=31.01\,\mathrm{g\,mol^{-1}}$,
$$
\boxed{m_{\rm CF}=\frac{630720}{96500}(31.01)=202.7\,\mathrm g}.
$$
This is active CF alone for the same single-cell-equivalent assumption, not the full positive-electrode or pack mass. Conductive additive, binder, <electrolyte>, <battery separator>, collectors and casing add mass, and incomplete utilization increases the required CF.
Back to article page
