a, PTHH: Ca+2H2O--->Ca(OH)2+H2 (1)

CaO+H2O--->Ca(OH)2 (2)

nH2= \(\dfrac{2,24}{22,4}=0,1\) mol

Theo pt(1) nH2=nCa=0,1 mol

=> mCa= 0,1.40= 4 g

mCaO= 17,2-4=13,2 g

=> %Ca= \(\dfrac{4}{17,2}.100\%\approx23,26\%\)

%CaO= 100%-23,26%= 76,74%

b, Theo pt(1) và (2): nCa(OH)2=nCaO= 0,1 mol

=> mCa(OH)2= 0,1.74= 7,4 g

M_KMnO4= 39+55+16.4 = 158 g

1152 đó bn tik MK ZỚI NHA!!!!!!!!!!!!

PTK_Al(NO3)3= 27+ (14+48).3= 213 đvC

M_Al(NO3)3= 213 g

PTK_Fe(OH)2= 56+ (16+1).2 = 90 đvC

M_Fe(OH)2= 90 g

PTHH: 2KOH+KH2PO4--->K3PO4+2H2O

Hoặc: KOH_(loãng)+KH2PO4---> K2HPO4+H2O

KOH_{(đậm đặc)}+K2HPO4---> K3PO4+H2O

a, Na Mg Mg Zn Cu Ag

b, PTHH: Mg+CuSO4--->MgSO4+Cu

Zn+CuSO4--->ZnSO4+Cu

a, PTHH: Fe+2HCl--->FeCl2+H2

nH2= \(\dfrac{0,224}{22,4}=0,01\) mol

Theo pt: nFe=nH2=0,01 mol

=> mFe= 0,01.56= 0,56 g

mCu= 1,2-0,56= 0,64 g

=> %Fe= \(\dfrac{0,56}{1,2}.100\%\approx46,7\%\)

%Cu= 100%-46,7%= 53,3%

b, Theo pt: nHCl=2.nH2= 2.0,01=0,02 mol

=> mHCl= 0,02.36,5= 0,73 g

=> C%ddHCl= \(\dfrac{0,73}{10}.100\%=7,3\%\)

a, PTHH: CuO+H2--->Cu+H2O

nCuO= \(\dfrac{9,6}{80}=0,12\) mol

Theo pt: nCu=nCuO= 0,12 mol

=> mCu= 0,12.64= 7,68 g

b, Theo pt: nH2=nCuO= 0,12 mol

=> VH2= 0,12.22,4= 2,688 l

có 2 kim loại hóa trị 2 nữa mà bn

m=n.M

n=\(\dfrac{m}{M}\)

M=\(\dfrac{m}{n}\)

V= n.22,4

n= \(\dfrac{V}{22,4}\)