= Solution
Each <lithium> atom supplies one <electron>. If all of the stated <lithium> is electrochemically accessible and there is enough <iron disulfide>, the ideal <electric charge> is
$$
Q=\frac{m_{\rm Li}}{M_{\rm Li}}F=\frac{0.453}{6.94}(96540)=\boxed{6.30\times10^3\,\mathrm C=1.75\,\mathrm{Ah}}.
$$
The <voltage> is not needed to calculate charge. At an assumed constant 1.5 V the corresponding ideal <energy> would be $VQ\simeq9.45\,\mathrm{kJ}=2.63\,\mathrm{Wh}$, but an actual discharge <voltage> varies with load and state of charge. Side reactions, incomplete utilization and the practical cutoff reduce accessible capacity. The preceding optimal ratio would require approximately 1.96 g of <iron disulfide> for this amount of <lithium>.
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