How to calculate the rate of formation of a product when given the rate law? The following data are given for the reaction of $\ce{NO} \text{ and } \ce{Cl2}$:

$$\ce{2NO + Cl2 -&gt; 2NOCl}$$
The reaction is second order in $[\ce{NO}]$ and first order in $[\ce{Cl2}]$,
and the  initial rate equals $\pu{1.43E-6 mol L^-1 s^-1}$ at the instant when $[\ce{NO}]_0 = [\ce{Cl2}]_0 = \pu{0.25 mol L^-1}$.

The problem then says:

Calculate the rate of formation of $\ce{NOCl}$ when $[\ce{NO}] = [\ce{Cl2}] = \pu{0.11 mol L^-1}$.

I thought it's just $(2) × (\pu{9.152E-5})× [0.11]^2[0.11]$, which is $\pu{2.44E-7}$,  but the answer provided is half that $(\pu{1.22E-7})$, so I must have gotten the rate equation incorrect.
Is the rate of formation of $\ce{NOCl}$ equal to twice the rate of the reaction, as when I consider this reaction equation: $\ce{2NO + Cl2 -&gt; 2NOCl}$?
Or is the rate of formation of  $\ce{NOCl}$ directly equal to the rate of the reaction, as when I consider this reaction equation: $\ce{NO + 1/2Cl2 -&gt; NOCl}$?
I'm really confused how to tell which of these two approaches to use.
$$r=k \times [NO_2]^2 \times [Cl_2]$$

From the initial condition information =>
$$k=\frac{r}{[NO_2]^2 \times [Cl_2]}$$
Plugging the numbers=>
$$k=\frac{1.43 \times 10^{-6}}{0.25^2 \times 0.25}$$
So, $$k=9.152 \times 10^{-5}$$
For 0.11 condition =>
$$r=9.152 \times 10^{-5} \times 0.11^2 \times 0.11$$
or $$r=1.21 \times 10^{-7}$$