a, PT: \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\)

b, Ta có: \(n_{H_2}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,5.65=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=72,5-32,5=40\left(g\right)\)

c, Ta có: \(n_{CuO}=\dfrac{40}{80}=0,5\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Zn}+n_{CuO}=1\left(mol\right)\)

\(\Rightarrow b=C_{M_{H_2SO_4}}=\dfrac{1}{2,5}=0,4M\)

c, Theo PT: \(\left\{{}\begin{matrix}n_{ZnSO_4}=n_{Zn}=0,5\left(mol\right)\\n_{CuSO_4}=n_{CuO}=0,5\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}C_{M_{ZnSO_4}}=\dfrac{0,5}{2,5}=0,2M\\C_{M_{CuSO_4}}=\dfrac{0,5}{2,5}=0,2M\end{matrix}\right.\)

Bạn tham khảo nhé!

a, \(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\left(I\right)\)

\(CuO+H_2SO_4\rightarrow CuSO_4+H_2O\left(II\right)\)

b, Theo PTHH(1) : \(n_{Zn}=n_{H_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{Zn}=32,5\left(g\right)\)

\(\Rightarrow m_{CuO}=m_{hh}-m_{Zn}=40\left(g\right)\)

\(\Rightarrow n_{CuO}=\dfrac{m}{M}=0,5\left(mol\right)\)

c, Theo PTHH (1) và (2) : \(n_{H2SO4}=n_{CuO}+n_{Zn}=1\left(mol\right)\)

\(\Rightarrow C_{MH2SO4}=b=\dfrac{n}{V}=\dfrac{1}{2,5}=0,4M\)

d, ( Chắc là thể tích coi như không đổi )

Thấy sau phản ứng thu được A gồm \(0,5molZnSO_4,0,5molCuSO_4\)

\(\Rightarrow C_{MCuSO4}=C_{MZnSO4}=\dfrac{n}{V}=\dfrac{0,5}{2,5}=0,2M\)

Vậy ...

Câu 3 : 

\(n_{HCl}=\dfrac{10\cdot21.9\%}{36.5}=0.06\left(mol\right)\)

\(AO+2HCl\rightarrow ACl_2+H_2O\)

\(0.03........0.06\)

\(M=\dfrac{2.4}{0.03}=80\left(\dfrac{g}{mol}\right)\)

\(\Rightarrow A=64\)

\(CuO\)

Câu 2 : 

$n_{CuO} = \dfrac{1,6}{80} = 0,02(mol)$
$n_{H_2SO_4} = \dfrac{100.20\%}{98} = \dfrac{10}{49}$

$CuO + H_2SO_4 \to CuSO_4 + H_2O$

$n_{CuO} < n_{H_2SO_4}$ nên $H_2SO_4 dư

Theo PTHH :
$n_{CuSO_4} = n_{H_2SO_4\ pư} = n_{CuO} = 0,02(mol)$
$m_{dd} = 1,6 + 100 = 101,6(gam)$

Vậy :

$C\%_{CuSO_4} = \dfrac{0,02.160}{101,6}.100\% = 3,15\%$

$C\%_{H_2SO_4\ dư} = \dfrac{100.20\% - 0,02.98}{101,6}.100\% = 17,6\%$

CÂU 2

m_H2SO4=100.20%=20(g)

n_H2SO4=20/98=0,2(mol)

n_CuO=1,6/80=0,02(mol)

PTHH : CuO+H₂SO₄-->CuSO₄+H₂O(1)

bài        0,02    0,2         0,02       0,02             (mol)

có:0,02/1<0,2/1---->CuO hết,H₂SO₄ dư

từ pt(1)-->n_CuSO4=0,02(mol)--->m_CuSO4=0,02.160=3,2(g)

khối lượng dd sau pư là:1,6+100-0,02.18=101,24(g)

-->C_%(CuSO4)=3,2/101,24.100%=3,16%

CÂU 3

m_HCl=10.21,9%=2,19(g)

-->n_HCl=2,19/36,5=0,06(mol)

gọi tên KL là M.M_M=M(g/mol)

PTHH: MO+2HCl-->MCl₂+H₂O(1)

          0,03     0,06                       (mol)

từ pt 1-->n_MO=0,03(mol)

--->M_MO=2,4/0,03=80(g/mol)

--->M=80-16=64(g/mol)

--->M là Cu

\(n_{Al}=\dfrac{0,54}{27}=0,02mol\\ n_{H_2SO_4}=0,07.0,5=0,035mol\\ 2Al+3H_2SO_4->Al_2\left(SO_4\right)_3+3H_2\\ \Rightarrow H_2SO_4:dư\\ V=\dfrac{3}{2}.0,02.22,4=0,672L\\ C_{M\left(H_2SO_4\right)}=\dfrac{0,005}{0,07}=0,071M\\ C_{M\left(Al_2\left(SO_4\right)_3\right)}=\dfrac{0,01}{0,07}=\dfrac{1}{7}\left(M\right)\\ CuO+H_2-t^{^0}->Cu+H_2O\\ n_{H_2}=0,03mol\\ n_{CuO}=\dfrac{6,4}{80}=0,08\Rightarrow CuO:dư\\ m_{rắn}=6,4-16.0,03=5,92g\)

\(a)Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_{\text{ 4}}\right)_3+3H_2O\\ n_{Fe_2O_3}=\dfrac{1,6}{160}=0,01\left(mol\right)\\ n_{H_2SO_4}=3n_{Fe_2O_3}=0,03\left(mol\right)\\m_{ddH_2SO_4}=\dfrac{0,03.98}{19,6\%}=15\left(g\right)\\ b)m_{ddsaupu}=1,6+15=16,6\left(g\right)\\ n_{Fe_2\left(SO_4\right)_3}=n_{Fe_2O_3}=0,01\left(mol\right)\\ C\%_{Fe_2\left(SO_4\right)_3}=\dfrac{0,01.400}{16,6}.100=24,1\%\)

\(n_{Fe2O3}=\dfrac{1,6}{160}=0,01\left(mol\right)\)

a) Pt : \(Fe_2O_3+3H_2SO_4\rightarrow Fe_2\left(SO_4\right)_3+3H_2O|\)

               1              3                   1                 3

             0,01         0,03                0,01

\(n_{H2SO4}=\dfrac{0,01.3}{1}=0,03\left(mol\right)\)

⇒ \(m_{H2SO4}=0,03.98=2,94\left(g\right)\)

\(m_{ddH2SO4}=\dfrac{2,94.100}{19,6}=15\left(g\right)\)

b) \(n_{Fe2\left(SO4\right)3}=\dfrac{0,03.1}{3}=0,01\left(mol\right)\)

⇒ \(m_{Fe2\left(SO4\right)3}=0,01.400=4\left(g\right)\)

\(m_{ddspu}=1,6+15=16,6\left(g\right)\)

\(C_{Fe2\left(SO4\right)3}=\dfrac{4.100}{16,6}=24,1\)⁰/₀

 Chúc bạn học tốt

a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

\(MgO+H_2SO_4\rightarrow MgSO_4+H_2O\)

b, \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

Theo PT: \(n_{Mg}=n_{H_2}=0,1\left(mol\right)\Rightarrow m_{Mg}=0,1.24=2,4\left(g\right)\)

\(\Rightarrow m_{MgO}=4,4-2,4=2\left(g\right)\)

c, \(n_{MgO}=\dfrac{2}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Mg}+n_{MgO}=0,15\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,15.98=14,7\left(g\right)\)

\(\Rightarrow m_{ddH_2SO_4}=\dfrac{14,7}{19,6\%}=75\left(g\right)\)

a) 4P + 5O₂ --t^o--> 2P₂O₅

b) \(n_P=\dfrac{15,5}{31}=0,5\left(mol\right)\)

PTHH: 4P + 5O₂ --t^o--> 2P₂O₅

           0,5--------------->0,25

=> m_P2O5 = 0,25.142 = 35,5 (g)

c) 

PTHH: P₂O₅ + 3H₂O --> 2H₃PO₄

              0,25------------>0,5

=> \(C\%=\dfrac{0,5.98}{100}.100\%=49\%\)

4P+5O2-to>2P2O5

0,5-------------0,25

n P=0,5 mol

=>m P2O5=0,25.142=35,5g

c)P2O5+3H2O->2H3PO4

    0,25---------------------0,5

=>C%=\(\dfrac{0,5.98}{100}100=49g\)

thôi thì mình làm cho bn vậy, câu a ko làm dc đâu, làm câu b thôi, làm sao biết dc chất nào dư khi chỉ có số mol 1 chất?

nK2SO3=0.1367(mol)

mddH2SO4=Vdd.D=200.1,04=208(g)

K2SO3+H2SO4-->K2SO4+H2O+SO2

0.1367----0.1367----0.1367---------0.1367   (mol)

mddspu=100+208-0,1367.64=299.2512(g) ; mK2SO4=0,1367.174=23.7858(g)

==>C%=23.7858.100/299.512=7.94%

2)pt bn tự ghi nhé

ta có hệ pt: 56a+27b=11 và a+3b/2=8.96/22.4==>a=0.1, b=0.2

==>%Fe=0.1x56x100/11=50.9%

%Al=100%-50.9%=49.1%

b)nH2SO4= 0.7(mol)==>VddH2SO4=0.7/2=0.35(L)

bn xem lại đề nhé, bài 1mình thấy số mol ra lẻ><

câu hỏi của bài 2 là j z bn?