-Gọi CTHH oxit: R2On
R2On+nH2SO4→→R2(SO4)n+nH2O
nH2SO4=7,84\98=0,08mol
nR2On=1\nH2SO4=0,08nmol
MR2On=4,48\0,08n=56n
2R+16n=56n→→2R=40n→→R=20n
n=1→R=20(loại)
n=2→R=40(Ca)
n=3→R=60(loại)
-CTHH oxit: CaO
a,\(M_xO_y+yH_2SO_4\rightarrow M_x\left(SO_4\right)_y+yH_2\)
Ta có :
\(n_{H2SO4}=\frac{7,84}{98}=0,08\left(mol\right)\)
\(\Rightarrow n_{MxOy}=\frac{0,08}{y}\left(mol\right)\)
\(M_{MxOy}=\frac{4,48}{\frac{0,08}{y}}=56y\)
\(\Rightarrow M_M=20.\frac{2y}{x}\)
\(\frac{2y}{x}=2\Rightarrow M_M=40\Rightarrow M:Ca\)
Vậy CTHH là CaO
b,
Cách 1 :
\(n_{CaO}=\frac{4,48}{56}=0,08\left(mol\right)\)
\(n_{CaSO4}=n_{CaO}=0,08\left(mol\right)\)
\(\Rightarrow m_{CaSO4}=0,08.136=10,88\left(g\right)\)
Cách 2 :
\(n_{H2O}=n_{H2SO4}=0,08\left(mol\right)\)
BTKL, \(m_{CaO}+m_{H2SO4}=m_{CaSO4}+m_{H2O}\)
\(\Rightarrow m_{CaSO4}=10,88\left(g\right)\)
