CuO +H2SO4---->CuSO4 +H2O
a) Ta có
n\(_{CuO}=\frac{10}{80}=0,125\left(mol\right)\)
Theo pthh
n\(_{CuSO4}=n_{CuO}=0,125\left(mol\right)\)
m\(_{CuSO4}=0,125.160=20\left(g\right)\)
b) Theo pthh
n\(_{H2SO4}=n_{CuO}=0,125\left(mol\right)\)
C\(_{M\left(H2SO4\right)}=\frac{0,125}{0,2}=0,625\left(M\right)\)
Chúc bạn học tốt
n
a, nCuO = 0,125 (mol)
PTHH : CuO + H2SO4 -->CuSO4+H2O
=> n CuSO4 = 0,125
=>mCuSO4 = 20(g)
b,CM H2SO4 = 0,125/0,2 = 0,625M