\(a,PTHH:2Cu+O_2\xrightarrow{t^o}2CuO\\ n_{Cu}=\dfrac{11,52}{64}=0,18(mol)\\ b,n_{CuO}=n_{Cu}=0,18(mol)\\ \Rightarrow m_{CuO}=0,18.80=14,4(g)\\ c,n_{O_2}=0,5.n_{Cu}=0,09(mol)\\ \Rightarrow V_{O_2}=0,09.22,4=2,016(l)\\ \Rightarrow V_{kk}=2,016.5=10,08(l)\)