\(n_{CuO}=\dfrac{15}{80}=0,1875\left(mol\right);n_{Cu}=\dfrac{13,6}{64}=0,2125\left(mol\right)\)
PTHH: CuO + H2 → Cu + H2O
Mol: 0,1875 0,1875
Tỉ lệ: \(\dfrac{0,1875}{1}< \dfrac{0,2125}{1}\)⇒ CuO hết, H2 dư
\(m_{H_2O}=0,1875.18=3,375\left(g\right)\)
\( Đặt:n_{CuO\left(bđ\right)}=\dfrac{15}{80}=0,1875\left(mol\right)\\ n_{CuO\left(p.ứ\right)}=a\left(mol\right)\\ CuO+H_2\underrightarrow{^{to}}Cu+H_2O\\ m_{rắn}=13,6\\ \Leftrightarrow\left(15-80a\right)+64a=13,6\\ \Leftrightarrow x=0,0875< 0,1875\\ \Rightarrow CuOdư\\ n_{Cu\left(p.ứ\right)}=0,1875-0,0875=0,1\left(mol\right)\\ n_{H_2O}=n_{Cu\left(p.ứ\right)}=0,1\left(mol\right)\\ m_{H_2O}=18.0,1=1,8\left(g\right)\)
