If \(q>0\) and \(i\in\{0,1,\dots\}\) then \([i]=(1-q^i)(1-q)^{-1}\) for \(q\neq 1\) and \([i]=i\) for \(q=1\), \([i]!=[i][i-1]\dots[1]\) for \(i\in \mathbb{N}\) and \([i]!=1\) for \(i=0\), \(\left[\begin{matrix} k \\ r\end{matrix}\right]=\left([k]!\right)\left([r]![k-r]!\right)^{-1}\). Let \(I^2=[0,1]\times[0,1]\) and \(\mathbb{R}^{I^2}=\{f:I^2 \rightarrow \mathbb{R}\}\). For \(f\in \mathbb{R}^{I^2}\), \(\{q_1,q_2\}\subset(0,\infty),\{n_1,n_2\}\subset \mathbb{N}\), the linear positive operator \(B_{n_1,n_2}(f;x,y)\) is \[ \sum^{n_1}_{r_1=0} \sum^{n_2}_{r_2=0} f\left(\frac{[r_1]}{[n_1]}, \frac{[r_2]}{[n_2]}\right)\left[\begin{matrix} n_1 \\ r_1\end{matrix}\right] \left[\begin{matrix} n_2 \\ r_2\end{matrix}\right] x^{r_1}y^{r_2} \prod^{n_1-r_1-1}_{s_1=0}(1-q^{s_1}_1 x)\prod^{n_2-r_2-1}_{s_2=0} (1-q^{s_2}_2 y). \] The author proves the following results: I) Let \(q_1=q_1(n_1)\), \(q_2=q_2(n_2)\) and let \(q_1(n_1)\to 1\), \(q_2(n_2)\to 1\) from below \(n_1,n_2\to\infty\). Then, for \(f\in C(I^2,\mathbb{R})\) the sequence \(B_{n_1,n_2}(f)\) converges uniformly to \(f\) on \(I^2\). II) For \(f:I^2\to \mathbb{R}\), bounded on \(I^2\), the inequality \[ \|f-B_{n_1,n_2}(f)\|_\infty \leq \left(\frac 32\right)^2 \omega_f \left([n_1]^{-0,5}, [n_2]^{-0,5}\right) \] holds, where \(\|\;.\|_\infty\) denotes the uniform norm and \(\omega_f\) denotes the first order modulus of smoothness.