\(m_{BaCl}=n\cdot M=0,2\cdot\left(137+35,5\right)=34,5\left(g\right)\)
\(n_{CO_2\left(dktc\right)}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ m_{CO_2}=n\cdot M=0,2\cdot44=8,8\left(g\right)\)
\(m_{CuO}=n\cdot M=0,25\cdot\left(64+16\right)=20\left(g\right)\)