Giải ra vì sao giúp tớ ạ,tớ cảm ơn

$28,A$

$n_{CaO}=0,1(mol)$

$CaO+2HCl \to CaCl_2+H_2O$

  $0,1$`->`$0,2(mol)$

`=>`$m_{HCl}=7,3(g)$

`=>`$m_{dd}=\dfrac{7,3 \cdot 100}{14,6}=50(g)$

$29,C$

$RO$`->`$\dfrac{R}{R+16} \cdot 100\%=71,43%$

`=>`$R=40$

`=>`$Ca$`=>`$CaO$

$30,$

`-` Bảo toàn khối lượng:

$m_{S}+m_{O_2}=m_{SO_2}$

`=>`$m_{O_2}=96-48=48(g)$

$31,C$

$n_{CaO} =\dfrac{5,6}{56}=0,1(kmol)$

$CaCO_3 \xrightarrow{t^o} CaO+CO_2$

  $0,1$               `←`               $0,1(kmol)$

`=>`$m_{CaCO_3}=0,1 \cdot 100 \cdot \dfrac{100}{95}=\dfrac{200}{19}(tấn)$

$32,B$

$m_{CaCO_3}=0,75(tấn)$

`=>`$n_{CaCO_3}=0,0075(kmol)$

$CaCO_3 \xrightarrow{t^o} CaO+CO_2$

   $0,0075$      `->`                $0,0075(kmol)$

`=>`$m_{CaO}=0,0075 \cdot 56 \cdot 90\%=0,378(tấn)=378(kg)$

$33,C$

$n_{CaO}=1(mol)$

$m_{CaO}=56(g)$

`->`$m_{H_2O}=80\% \cdot 56=44,8(g)$

$CaO+H_2O \to Ca(OH)_2$

`->`$n_{H_2O}=n_{CaO}=(mol)$

`=>`$m_{H_2O}=18(g)$

`=>`$44,8:18=\dfrac{112}{45}$

$34,D$

$n_{SO_3}=0,15(mol)$

$SO_3+H_2O \to H_2SO_4$

`=>`$C_M(H_2SO_4)=\dfrac{0,15}{0,25}=0,6(M)$

$35,A$

$n_{SO_2}=0,05(mol)$

$n_{Ca(OH)_2}=0,06(mol)$

$T=\dfrac{0,06 \cdot 2}{0,05}=2,4$

`=>`$CaSO_3$ tính theo $SO_2.$

$Ca(OH)_2+SO_2 \to CaSO_3+H_2O$

`=>`$n_{CaSO_3}=0,05(mol)$

`=>`$m_{CaSO_3}=0,05 \cdot 120=6(g)$