\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

\(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(NO\) là sản phẩm khử duy nhất.

\(\Rightarrow n_{NO}=\dfrac{0,448}{22,4}=0,02mol\)

Ta có: \(\left\{{}\begin{matrix}27n_{Al}+108n_{Ag}=3,51g\\BTe:3n_{Al}+n_{Ag}=3n_{NO}=0,06\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,01mol\\n_{Ag}=0,03mol\end{matrix}\right.\)

\(\%m_{Al}=\dfrac{0,01\cdot27}{3,51}\cdot100\%=7,7\%\)

\(\%m_{Ag}=100\%-7,7\%=92,3\%\)

a, \(Al+4HNO_3\rightarrow Al\left(NO_3\right)_3+NO+2H_2O\)

\(3Ag+4HNO_3\rightarrow3AgNO_3+NO+2H_2O\)

b, \(\left\{{}\begin{matrix}\underrightarrow{BTE}3n_{Al}+n_{Ag}=\dfrac{0,448.3}{22,4}=0,06\\\underrightarrow{BTKL}27n_{Al}+108n_{Ag}=3,51\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}n_{Al}=0,01\\n_{Ag}=0,03\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\%Al=\dfrac{0,01.27}{3,51}.100\%=7,69\%\\\%Ag=\dfrac{0,03.108}{3,51}.100\%=92,31\%\end{matrix}\right.\)

Đề này có cho D của dung dịch ko em?

Sửa đề : 200ml thành 200g 

a) \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(3H_2+Fe_2O_3-^{t^o}\rightarrow2Fe+3H_2O\) (2)

b) 1/2 lượng khí B: \(n_{H_2\left(2\right)}=3n_{Fe_2O_3}=3.\dfrac{38,4}{160}=0,72\left(mol\right)\)

=> \(n_{H_2\left(1\right)}=0,72.2=1,44\left(mol\right)\)

\(n_{H_2SO_4}=n_{H_2\left(2\right)}=1,44\left(mol\right)\)

=> \(C\%H_2SO_4=\dfrac{1,44.98}{200}.100=70,56\%\)

\(n_{Al}=\dfrac{2}{3}n_{H_2\left(2\right)}=0,96\left(mol\right)\)

=> \(m_{Al}=0,96.27=25,92\left(g\right)\)

Có lẽ đề bài cho 200 g dd H₂SO₄ chứ không phải 200 ml bạn nhỉ?

a, PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\) (1)

\(FeO+H_2\underrightarrow{t^o}Fe+H_2O\) (2)

b, Ta có: \(n_{FeO}=\dfrac{38,4}{72}=\dfrac{8}{15}\left(mol\right)\)

Theo PT: \(n_{H_2\left(2\right)}=n_{FeO}=\dfrac{8}{15}\left(mol\right)\)

\(\Rightarrow n_{H_2\left(1\right)}=\dfrac{8}{15}.2=\dfrac{16}{15}\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{32}{45}\left(mol\right)\\n_{H_2SO_4}=n_{H_2}=\dfrac{16}{15}\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow x=m_{Al}=\dfrac{32}{45}.27=19,2\left(g\right)\)

\(m_{H_2SO_4}=\dfrac{16}{15}.98=\dfrac{1568}{15}\left(g\right)\)

\(\Rightarrow y=C\%_{H_2SO_4}=\dfrac{\dfrac{1568}{15}}{200}.100\%\approx52,27\%\)

Bạn tham khảo nhé!

600ml = 0,6l

600ml = 0,6l

\(n_{H2SO4}=1.0,6=0,6\left(mol\right)\)

\(n_{BaCl2}=0,5.0,6=0,3\left(mol\right)\)

Pt : \(H_2SO_4+BaCl_2\rightarrow2HCl+BaSO_4|\)

           1             1              2             1

         0,6           0,3          0,6           0,3

a) Lập tỉ số so sánh : \(\dfrac{0,6}{1}>\dfrac{0,3}{1}\)

             ⇒ H₂SO₄ dư , BaCl₂ phản ứng hết

             ⇒ Tính toán dựa vào số mol của BaCl₂

\(n_{BaSO4}=\dfrac{0,3.1}{1}=0,3\left(mol\right)\)

⇒ \(m_{BaSO4}=0,3.233=69,9\left(g\right)\)

b) \(n_{HCl}=\dfrac{0,3.2}{1}=0,6\left(mol\right)\)

 \(n_{H2SO4\left(dư\right)}=0,6-0,3=0,3\left(mol\right)\)

\(V_{ddspu}=0,6+0,6=1,2\left(l\right)\)

\(C_{M_{HCl}}=\dfrac{0,6}{1,2}=0,5\left(M\right)\)

\(C_{M_{H2SO4\left(dư\right)}}=\dfrac{0,3}{1,2}=0,25\left(M\right)\)

 Chúc bạn học tốt

a) nMg= 2,4/24=0,1(mol); nAl=5,4/27=0,2(mol)

PTHH: Mg + H2SO4 -> MgSO4 + H2

0,1__________0,1_____0,1____0,1(mol)

PTHH: 2Al + 3 H2SO4 -> Al2(SO4)3 +3 H2

0,2_________0,3_______0,1________0,3(mol)

nH2SO4(tổng)=nH2(tổng)=0,1+0,3=0,4(mol)

V(H2,đktc)=(0,1+0,3).22,4=8,96(l)

b) mH2SO4=39,2(g)

CMddH2SO4=0,3/0,1=3(M)

=> C%ddH2SO4= (C_MddH2SO4 .M(H2SO4) ) /(10D)= (3.98)/(10.1,2)=24,5% 

Chúc em học tốt!

a) PTHH: \(FeCl_3+3NaOH\rightarrow3NaCl+Fe\left(OH\right)_3\downarrow\)

                 \(2Fe\left(OH\right)_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+6H_2O\)

b) Ta có: \(n_{FeCl_3}=0,3\cdot0,5=0,15\left(mol\right)\)

\(\Rightarrow n_{NaOH}=0,45mol\) \(\Rightarrow V_{ddNaOH}=\dfrac{0,45}{0,25}=1,8\left(l\right)\)

c) Theo PTHH: \(n_{NaCl}=n_{NaOH}=0,45mol\)

\(\Rightarrow C_{M_{NaCl}}=\dfrac{0,45}{2,1}\approx0,21\left(M\right)\) 

(Coi như thể tích dd thay đổi không đáng kể)

d) Theo PTHH: \(n_{H_2SO_4}=\dfrac{3}{2}n_{Fe\left(OH\right)_3}=\dfrac{3}{2}n_{FeCl_3}=0,225mol\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{0,225\cdot98}{20\%}=110,25\left(g\right)\) 

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{110,25}{1,14}\approx96,71\left(ml\right)\)

\(3NaOH+FeCl_3\rightarrow Fe\left(OH\right)_3+3NaCl\)

\(n_{NaCl}=n_{NaOH}=0,2.3=0,6\left(mol\right)\)

=> \(C_{M\left(NaCl\right)}=\dfrac{0,6}{0,2}=3M\)

\(n_{Fe\left(ỌH\right)_3}=\dfrac{1}{3}n_{NaOH}=0,2\left(mol\right)\)

\(2Fe\left(OH\right)_3-^{t^o}\rightarrow Fe_2O_3+3H_2O\)

Ta có \(n_{Fe_2O_3}=\dfrac{1}{2}n_{Fe\left(OH\right)_3}=0,1\left(mol\right)\)

=> m _Fe2O3 = 0,1 . 160=16(g)