\(n_{Cu}=\dfrac{6,4}{64}=0,1\left(mol\right)\)
Pt : \(2Cu+O_2\underrightarrow{t^o}2CuO|\)
2 1 2
0,1 0,05 0,1
a) \(n_{O2}=\dfrac{0,1.1}{2}=0,05\left(mol\right)\)
\(V_{O2\left(dktc\right)}=0,05.22,4=1,12\left(l\right)\)
b) \(n_{CuO}=\dfrac{0,05.2}{1}=0,1\left(mol\right)\)
⇒ \(m_{CuO}=0,1.80=8\left(g\right)\)
Chúc bạn học tốt
\(Cu+\dfrac{1}{2}O_2\underrightarrow{t^o}CuO\)
\(n_{Cu}=\dfrac{m}{M}=\dfrac{6,4}{64}=0,1mol\)
\(V_{O_2}=n.22,4=0,1.\dfrac{1}{2}.22,4=1,12l\)
\(m_{sp}=n.M=0,1.80=8g\)