Given 7 equiv calcium atoms; 10 equiv of carbon atoms; and 9 equiv of nitrogen atoms, how many equiv of "calcium cyanide" could we make?

The formula for $\text{calcium cyanide}$ is $C a {\left(C \equiv N\right)}_{2}$; we could make four such formula units with the given quantities.
$\text{Calcium cyanide}$ is formulated by taking the calcium ion, $C {a}^{2 +}$, and binding it to 2 equiv of the cyanide ion, ""^(-):C-=N.
So for $7 \times C a$, $10 \times C$, and $9 \times N$, at most we could make $9 {\times}^{-} : C \equiv N$ anions.
However from these anions, we could make AT MOST only $4 \times C a {\left(C \equiv N\right)}_{2}$ units. Are you happy with this?