\(n_{CuO}=\dfrac{12}{80}=0,15mol\)
\(n_{H_2SO_4}=\dfrac{78,4.25\%}{98}=0,2mol\)
\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)
0,15 < 0,2 ( mol )
0,15 0,15 0,15 ( mol )
\(m_{CuSO_4}=0,15.160=24g\)
\(m_{H_2SO_4\left(dư\right)}=\left(0,2-0,15\right).98=4,9g\)
\(m_{ddspứ}=12+78,4=90,4g\)
\(C\%_{CuSO_4}=\dfrac{24}{90,4}.100\%=26,54\%\)
\(C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{90,4}.100\%=5,42\%\)
a) \(n_{CuO}=\dfrac{12}{80}=0,15\left(mol\right)\)
\(m_{H_2SO_4}=78,4.25\%=19,6\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{19,6}{98}=0,2\left(mol\right)\)
Ta có: \(\dfrac{0,15}{1}< \dfrac{0,2}{1}\) ⇒ CuO hết, H2SO4 dư
CuO + H2SO4 -------> CuSO4 + H2O
0,15 0,15 0,15
\(C\%_{ddCuSO_4}=\dfrac{0,15.160.100\%}{12+78,4}=36,55\%\)
\(C\%_{ddH_2SO_4dư}=\dfrac{\left(0,2-0,15\right).98.100\%}{12+78,4}=5,42\%\)
CuO + H2SO4 --> CúO4 + H2O
a, nCu=0,1875 mol =nCuSO4 -->mCuSO4=30g
b, %CúSO4=mCUSO4/mdd=30/(12+78,4*25%)=94,93%
