1/
a,
Gọi số mol $KHSO_3: x$
$K_2CO_3:y$
$KHSO_3 +HCl \to KCl+SO_2+H_2O$
x x x x
$K_2CO_3+2HCl \to 2KCl+CO_2+H_2O$
y 2y 2y y
$mHCl=400.7,3\%=29,2g$
$nHCl\text{ban đầu}=\frac{29,2}{36,5}=0,8$
Ta có $d\frac{X}{H_2}=25,33$
⇒$MX=50,66$
⇒$\frac{64x+44y}{x+y}=50,66$
⇔$13,34x-6,66y=0(1)$
Ta có $mKHSO_3+mK_2CO_3=39,6$
⇒$136x+138y=39,6(2)$
(1)(2)⇒$\left \{ {{nKHSO_3=0,096} \atop {nK_2CO_3=0,19}} \right.$
$mHCl\text{phản ứng}=x+2y=0,096+0,19.2=0,476$
Ta có $mHCl\text{phản ứng}<nHCl\text{ban đầu}$
⇒Có axit dư
b,
$nHCl dư=0,8-0,476=0,324$
$mdd sau=39,6+400-0,096.64-0,19-44≈425,1g$
$C\%KCl=\frac{74,5.(0,096+0,19.2)}{425,1}.100=8,34\%$
$C\%HCl dư=\frac{0,324.36,5}{425,1}.100=2,78\%$
2/
Gọi số mol $Fe_2O_3:x$
$MgO:y$
$Al_2O_3:z$
$Fe_2O_3+3CO \to 2Fe+3CO_2$
x 2x
$Fe+2HCl \to FeCl_2+H_2$
2x 4x 2x
$MgO+2HCl \to MgCl_2+H_2O$
y 2y y
$Al_2O_3+6HCl \to 2AlCl_3+3H_2O$
z 6z 2z
Ta có $mFe_2O_3+mMgO+mAl_2O_3=2,22$
⇒$160x+40y+102z=2,22(1)$
Ta có $mFe+mMgO+mAl_2O_3=1,98$
⇒$2.56x+40y+102z=1,98(2)$
$100ml=0,1lit$
Ta có: $nHCl=0,1.1=0,1$
⇒$4x+2y+6z=0,1(3)$
(1)(2)(3)⇒$x=0,005 ; y=0,01 ; z=0,01$
$\%mFe_2O_3=\frac{0,05.160}{2,22}.100=36,04\%$
$\%mMgO=\frac{0,01.40}{2,22}.100=18,02\%$
$\%mAl_2O_3=100-36,04-18,02=45,94\%$
3/
Gỉa sử có 100gdd $H_2SO_4$ và gọi phần trăm $H_2SO_4$ là x
⇒$mH_2=100.5\%=5g$
$nH_2=\frac{5}{2}=2,5$
Ta có $mH_2SO_4=100.x\%$
⇒$mH_2O=100-100.x\%$
$nH_2O=\frac{100-100.x\%}{18}$
$nH_2SO_4=\frac{100.x\%}{98}$
$2Na+H_2SO_4 \to Na_2SO_4+H_2$
$Mg+H_2SO_4 \to MgSO_4+H_2$
$Na+H_2O \to NaOH +1/2H_2$
Theo pt ta có
$nH_2=nH_2SO_4+1/2nH_2O$
⇔$2,5=\frac{100.x\%}{98}+1/2.\frac{100-100.x\%}{18}$
⇒$x=15,8\%$
⇒$C\%H_2SO_4=15,8\%$
