\(a.n_{CuO}=\dfrac{16}{80}=0,2\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ 0,2......0,2............0,2......0,2\left(mol\right)\\ m_{ddH_2SO_4}=\dfrac{0,2.98.100}{10}=196\left(g\right)\\ m_{ddCuSO_4}=196+16=212\left(g\right)\\ C\%_{ddCuSO_4}=\dfrac{0,2.160}{212}.100\approx15,094\%\)
\(b.2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ 0,4........0,2..............0,2...............0,4\left(mol\right)\\ V_{ddNaOH}=\dfrac{0,4}{2}=0,2\left(l\right)=200\left(ml\right)\)
Đúng 2
Bình luận (0)
