a, Ta có: \(n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\)
PT: \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(n_{Cu}=n_{CuO}=0,2\left(mol\right)\Rightarrow m_{Cu}=0,2.64=12,8\left(g\right)\)
b, \(n_{H_2}=n_{CuO}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)
\(n_{CuO}=\dfrac{m}{M}=\dfrac{16}{64+16}=0,2\left(mol\right)\)
\(PTHH:Cu+H_2O\rightarrow CuO+H_2\)
1 1 1 1
0,2 0,2 0,2 0,2
a) \(m_{Cu}=n.M=0,2.64=12,8\left(g\right)\)
b) \(V_{H_2}=n.24,79=4,958\left(l\right).\)
n\(_{CuO}\)=\(\dfrac{16}{80}\)=0,2(m)
PTHH:CuO+H\(_2\)->Cu+H\(_2\)O
tỉ lệ :1 1 1 1
số mol:0,2 0,2 0,2 0,2
a)m\(_{Cu}\)=0,2.64=12,8(g)
b)V\(_{H_2}\)=0,2.22,4=44,8(l)
