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Z tác dụng với $H_2SO_4$ tạo khí suy ra Z có Al,Cu
$2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2$
$n_{Al} = \dfrac{2}{3}n_{H_2} = 0,02(mol)$
$m_{Cu} = 2,46 - 0,02.27=1,92(gam)$
$n_{Al(OH)_3} = 0,07(mol) ; n_{HCl} = 0,11(mol)$
NaAlO2 + HCl + H2O → Al(OH)3 + NaCl
a................a........................a...............................(mol)
Al(OH)3 + 3HCl → AlCl3 + 3H2O
a - 0,07......3a-0,21..............................(mol)
Suy ra: a + 3a - 0,21 = 0,11
Suy ra: a = 0,08
$2Na + 2H_2O \to 2NaOH + H_2$
$Na_2O + H_2O \to 2NaOH$
$2Al + 2NaOH + 2H_2O \to 2NaAlO_2 + 3H_2$
Ta có : $n_{H_2} = 0,5n_{Na} + 1,5n_{NaAlO_2} \Rightarrow n_{Na} = \dfrac{0,135 - 1,5.0,08}{0,5} = 0,03(mol)$
Bảo toàn Na,Al
$n_{Na_2O} = \dfrac{0,08 - 0,03}{2} = 0,025(mol)$
$n_{Al} = 0,08 + 0,02 = 0,1(mol)$
$\%m_{Al} = \dfrac{0,1.27}{0,1.27 + 0,025.62 + 0,03.23 + 1,92}.100\% = 39,36\%$
a) Đặt: nZn=x(mol); nFe= y(mol) (x,y: nguyên, dương)
Zn + H2SO4 -> ZnSO4 + H2
x_______x_______x________x
Fe + H2SO4 -> FeSO4 + H2
y____y_________y___y(mol)
b) m(rắn)=mCu=3(g)
=> m(Zn, Fe)= 21,6 - 3= 18,6(g)
Ta có hpt:
\(\left\{{}\begin{matrix}65x+56y=18,6\\22,4x+22,4y=6,72\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,2\\y=0,1\end{matrix}\right.\)
=> Zn= 65.0,2=13(g)
=>%mZn= (13/21,6).100=60,185%
%mCu=(3/21,6).100=13,889%
=>%mFe=25,926%
c) nH2SO4=x+y=0,3(mol) =>mH2SO4=29,4(g)
=> mddH2SO4= (29,4.100)/25=117,6(g)
Theo bài ra, ta có: \(m_{Ag}=5,6\left(g\right)\)
PTHH: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\uparrow\)
a) Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\) \(\Rightarrow n_{Al}=\dfrac{1}{15}\left(mol\right)\) \(\Rightarrow m_{Al}=\dfrac{1}{15}\cdot27=1,8\left(g\right)\)
\(\Rightarrow\%m_{Al}=\dfrac{1,8}{1,8+5,6}\cdot100\%\approx24,32\%\) \(\Rightarrow\%m_{Ag}=75,68\%\)
b) Theo PTHH: \(n_{H_2SO_4}=n_{H_2}=0,1mol\) \(\Rightarrow C_{M_{H_2SO_4}}=\dfrac{0,1}{0,1}=1\left(M\right)\)
c) PTHH: \(H_2SO_4+Ba\left(OH\right)_2\rightarrow BaSO_4\downarrow+2H_2O\)
Theo PTHH: \(n_{Ba\left(OH\right)_2}=n_{H_2SO_4}=0,1mol\)
\(\Rightarrow V_{ddBa\left(OH\right)_2}=\dfrac{0,1}{0,2}=0,5\left(l\right)=500\left(ml\right)\)
1,spư Z ko tan trong H2SO4 loang=>z la Cu=>Zn tan het tao Zn2+ VA TOAN BO Fe(3+) len Fe2+=>chat ma pư vs KMnO4 la Fe2+=>nFe2+=nFe3+=0.25mol
pt 5Fe2(+)+MnO4(-)+8H(+)=5Fe(3+)+Mn(2+)+4H2...
0.25----->0.05
Cm KMnO4=a=0.05/0.2=0.25M
2,
2,nFe2(SO4)3=0.25*1=0.25mol=>nFe3+=0.5 mol
spư thu dc ran z ma ko tan trong H2SO4 loang=>ran Z la Cu,nCu dư=3.28/64=0.05125mol=>Fe tan het
va toan bo Fe(3+) chuyen len het Fe(2+) do Cu dư
dat nFe=x,nCu pư=y=>56x+64y+3.28=17.8 (1)
mat khac theo bt e
Fe=Fe(2+)+2e................Fe(3+)+1e=F...
x-------------> 2x 0.5---->0.5
Cu=Cu(2+)+2e
y------------->2y
vi ne cho=ne nhan=>2x+2y=0.5(2)=>x=0.185,y=0.065
mCu=3.28+0.065*64=7.44g
3,
nMg=0.02mol,nFe=0.03mol
gia sư pư dung lai o b1
Mg+Cu(2+)=Mg(2+)+Cu
0.02--------------------->0.02
=>mX=mFe chua pư+mCu=2.96(g)
gia su pư dung lai o b2 tu toan bo Fe va Mg pư het vs Cu2+
Mg+Cu(2+)=Mg(2+)+Cu
0.02--------------------->0.02
Fe+Cu(2+)=Fe(2+)+Cu
0.03--------------------->0.03=>mX=(0.0...
do 2.96<mX=3.12<3.2 nen pư xay ra o TH het Mg,Fe pư nhung con dư
dat ,nFe pư=y,nFedư=0.03-y
Mg+Cu(2+)=Mg(2+)+Cu
0.02---------------------->0.02
Fe+Cu(2+)=Fe(2+)+Cu
y-------------------------->y
ta co mX=3,12=mCu tt +mFe dư=64(0.02+y)+56(0.03-y)=>y=0.02
=>nCuCl2=nCu(2+)=0.04 mol