a,\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\)
PTHH: CuO + H2SO4 →CuSO4 + H2O
Mol: 0,25 0,25 0,25
\(m_{ddH_2SO_4}=\dfrac{0,25.98.100}{19,6}=125\left(g\right)\)
b,mdd sau pứ = 20+125 = 145 (g)
\(C\%_{ddCuSO_4}=\dfrac{0,25.160.100\%}{145}=27,59\%\)
\(Cu+H_2SO_4\rightarrow CuSO_4+H_2\)
0,3125 0,3125 0,3125 (mol)
a)\(n_{Cu}=\dfrac{20}{64}=0,3125\left(mol\right)\)
\(m_{H_2SO_4}=0,3125.98=30,625\left(g\right)\)
\(m_{ddH_2SO_4}=\dfrac{30,625}{19,6}.100=156,25\left(g\right)\)
b)\(m_{CuSO_4}=0,3125.160=50\left(g\right)\)
\(m_{ddCuSO_4}=20+156,25=176,25\left(g\right)\)
\(C\%_{CuSO_4}=\dfrac{50}{176,25}.100\approx28,37\%\)
\(n_{CuO}=\dfrac{20}{80}=0,25\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ a.0,25.........0,25..........0,25.......0,25\left(mol\right)\\ Đặt:a=m_{ddH_2SO_4}\left(g\right)\Rightarrow m_{ddX}=a+20\left(g\right)\\Vì.axit.dư\Rightarrow m_{CuSO_4}=0,25.160=40\left(g\right)\\ \dfrac{40}{a+20}.100\%=16\%\Leftrightarrow a=230\left(g\right)\\ \Rightarrow m_{ddH_2SO_4}=230\left(g\right)\\ b.m_{H_2SO_4\left(dư\right)}=230.19,6\%-0,25.98=20,58\left(g\right)\\ C\%_{ddH_2SO_4\left(dư\right)}=\dfrac{20,58}{230+20}.100=8,232\%\)
