nCuO = 1.6/80=0.02 mol
CuO + 2HCl --> CuCl2 + H2O
0.02___0.04_____0.02
mCuCl2= 0.02*135=2.7 g
mHCl = 0.04*36.5=1.46g
Sửa đề:
Cho 1,6g đồng 2 hidric => đồng 2 oxit tác dụng HCl
a) \(n_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
PTHH: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)
Theo PTHH: \(n_{CuO}:n_{CuCl_2}=1:1\)
\(\Rightarrow n_{CuCl_2}=n_{CuO}=0,02\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,02.135=2,7\left(g\right)\)
b) Theo PTHH: \(n_{CuO}:n_{HCl}=1:2\)
\(\Rightarrow n_{HCl}=n_{CuO}.2=0,02.2=0,04\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,04.36,5=1,46\left(g\right)\)
CuO + 2HCl → CuCl2 + H2O
\(n_{CuO}=\frac{1,6}{80}=0,02\left(mol\right)\)
a) theo pt: \(n_{CuCl_2}=n_{CuO}=0,02\left(mol\right)\)
\(\Rightarrow m_{CuCl_2}=0,02\times135=2,7\left(g\right)\)
b) Theo pT: \(n_{HCl}=2n_{CuO}=2\times0,02=0,04\left(mol\right)\)
\(\Rightarrow m_{HCl}=0,04\times36,5=1,46\left(g\right)\)
