\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(n_{H_2SO_4}=n_{MgO}=0,02\left(mol\right)\)
\(\Rightarrow C\%_{H_2SO_4}=\dfrac{0,02.98}{50}.100\%=3,92\%\)
\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)
PTHH :
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
0,02 0,02
\(a,m_{H_2SO_4}=0,02.98=1,96\left(g\right)\)
\(C\%=\dfrac{1,96}{50}.100\%=3,92\%\)
\(n_{MgO}=\dfrac{0,8}{40}=0,02\left(mol\right)\)
\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)
\(n_{MgO}=n_{H2SO4}=0,02\left(mol\right)\)
\(C\%_{ddH2SO4}=\dfrac{0,02.98}{50}.100\%=3,92\%\)
\(n_{MgO}=\dfrac{0,8}{40}=0,02mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{H_2SO_4}=n_{MgO}=0,02mol\\ C_{\%H_2SO_4}=\dfrac{0,02.98}{50}\cdot100=3,92\%\)
