There is no way to produce $11.62 \pu{g}$ of $\ce{CaO}$ from a $\ce{CaCl2 + NaCl}$ sample weighing $110.0 \pu{g}$ which is $15.2\%$ $\ce{CaCl2}$ by weight. Your calculations are correct.*
Assuming $100\%$ yield, and, as mentioned, no hydration whatsoever, you need at least $20.96\%$ $\ce{CaCl2}$ in a $110.0 \pu{g}$ $\ce{CaCl2 + NaCl}$ sample to produce $11.62 \pu{g}$ of $\ce{CaO}$.
*However, if the final product is hydrated, this is possible. Let me know if you need help with calculating percentage hydration.
Underlying Principle
This is an application of the well-known law of conservation of matter, also known as the principle of atom conservation (POAC). You may refer to R. C. Mukherjee (2004), Modern Approach to Chemical Calculations: An Introduction to the Mole Concept, 7$^\text{th}$ edition for a thorough reading with lots of practice problems.
Calculation
Your overall (contracted) reaction is:
$$
\ce{CaCl2 -> -> CaO}
$$
We simply apply POAC:
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} \tag{1}
$$
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaO} = \dfrac{m_\ce{CaO}}{M_\ce{CaO}}M_\ce{Ca} = \dfrac{11.62}{56.077}40.078 \pu{g} = 8.30\pu {g} \tag{2}
$$
$$
\text{moles of }\ce{Ca}\text{ in }\ce{CaCl2} = \dfrac{m_\ce{CaCl2}}{M_\ce{CaCl2}}M_\ce{Ca} = \dfrac{m_\ce{CaCl2}}{110.98}40.078 = 0.36m_\ce{CaCl2} \tag{3}
$$
where, in Equations (2) and (3), $M_i$ is the molar mass of $i$ and $m_i$ is the mass of $i$ in sample of.
Equating Equations (2) and (3) in accordance with Equation (1):
$$
8.30\pu {g} = 0.36m_\ce{CaCl2}
\implies m_\ce{CaCl2} = 23.06 \pu{g}
$$
Thus, if the original $\ce{CaCl2 + NaCl}$ sample weighed $110.0 \pu{g}$, $\ce{CaCl2}$ was present in $\dfrac{23.06}{110.0}\times100 \%$ weight ratio, which is $20.96\%$
Note: All calculations have been rounded off to the second decimal place.
