\(n_{Cu}=\dfrac{6,4}{64}=0,1mol\)
\(m_{Cl}=13,5-6,4=7,1gam\)
\(n_{Cl}=\dfrac{7,1}{35,5}=0,2mol\)
-Gọi công thức là CuxCly
\(\dfrac{x}{y}=\dfrac{n_{Cu}}{n_{Cl}}=\dfrac{0,1}{0,2}=\dfrac{1}{2}\rightarrow CuCl_2\)
Cu+Cl2\(\overset{t^0}{\rightarrow}\)CuCl2
\(n_{Cl_2}=n_{Cu}=0,1mol\rightarrow V_{Cl_2}=0,1x22,4=2,24l\)