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

a, Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(n_{H_2SO_4}=0,2.1=0,2\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\). ta được H₂SO₄ dư.

Theo PT: \(n_{H_2SO_4\left(pư\right)}=n_{ZnSO_4}=n_{Zn}=0,1\left(mol\right)\)

\(\Rightarrow n_{H_2SO_4\left(dư\right)}=0,2-0,1=0,1\left(mol\right)\)

b, Ta có: \(m_{ZnSO_4}=0,1.161=16,1\left(g\right)\)

c, \(C_{M_{H_2SO_4\left(dư\right)}}=\dfrac{0,1}{0,2}=0,5M\)

\(C_{M_{ZnSO_4}}=\dfrac{0,1}{0,2}=0,5M\)

Bạn tham khảo nhé!

$n_{HCl} = 0,4(mol)$
Ba(OH)₂ +  2HCl → BaCl₂ + 2H₂O

0,2................0,4.........0,2.........................(mol)

$x = \dfrac{0,4}{0,2} = 2M$

$V_{dd} = 0,4 + 0,2 = 0,6(mol)$
$C_{M_{BaCl_2}} = \dfrac{0,2}{0,6} = 0,33M$

n _{CuSO4 (bđ)} = 0,4 . 0,5 = 0,2 (mol)

2Al + 3CuSO₄  Al₂(SO₄)₃ + 3Cu

Cứ 1 mol Cu tạo thành thì khối lượng CR tăng 138g 

       0,01 mol                                                   (11,38 - 10 ) g

m_Cu= 0,01 . 64 = 0,64 (g)

2Al + 3CuSO₄  Al₂(SO₄)₃ + 3Cu

               0,01               0,01    (mol)

Sau  pư thể tích dd ko đổi => V = 0,5 l

C_M(CuSO4)= = 0,38 (M)

C_M(Al2(SO4)3)=  = 0,007(M)

a) Số mol CuO và H2SO4 lần lượt là:

nCuO =\(\dfrac{16}{80}\)= 0,02 (mol)

nH2SO4 = \(\dfrac{100.20}{100.98}\) ≃ 0,2 (mol)

Tỉ lệ: nCuO : nH2SO4 = 0,02/1 : 0,2/1 = 0,02 : 0,1

=> H2SO4 dư, tính theo CuO

=> mH2SO4(dư) = n(dư).M = 0,18.98 = 17,64 (g)

PTHH:CuO + H2SO4 -> CuSO4 + H2O

---------0,02-------0,02------0,02----0,02--

=> Có 1,6 g CuO tham gia phản ứng

b) Khối lượng CuSO4 tạo thành là:

mCuSO4 = n.M = 0,02.160 = 3,2 (g)

c) Khối lượng dd sau phản ứng là:

mddspư = mCuO + mddH2SO4 = 1,6 + 100 = 101,6 (g)

Nồng độ phần trăm của axit thu được spư là:

C%H2SO4(dư) = \(\dfrac{17,64}{101,6}\) .100≃ 17,4 %

a) \(n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\)

\(n_{H_2SO_4}=\dfrac{20\%.100}{98}=\dfrac{10}{49}\left(mol\right)\)

PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O

Xét tỉ lệ: \(\dfrac{0,02}{1}< \dfrac{\dfrac{10}{49}}{1}\) => CuO hết, H₂SO₄ dư

b)

PTHH: CuO + H₂SO₄ --> CuSO₄ + H₂O

            0,02-->0,02------->0,02

=> \(m_{CuSO_4}=0,02.160=3,2\left(g\right)\)

c) 

m_{dd sau pư} = 1,6 + 100 = 101,6 (g)

\(C\%_{H_2SO_4.dư}=\dfrac{98.\left(\dfrac{10}{49}-0,02\right)}{101,6}.100\%=17,756\%\)

\(a.n_{CuO}=\dfrac{1,6}{80}=0,02\left(mol\right)\\ n_{H_2SO_4}=\dfrac{100.20\%}{98}=\dfrac{10}{49}\left(mol\right)\\ CuO+H_2SO_4\rightarrow CuSO_4+H_2O\\ LTL:\dfrac{0,02}{1}< \dfrac{10}{49}\Rightarrow H_2SO_4dư\\ b.n_{CuSO_4}=n_{Cu}=0,02\left(mol\right)\\ \Rightarrow m_{CuSO_4SS}=0,02.160=3,2\left(g\right)\\ c.m_{ddsaupu}=1,6+100=101,6\left(g\right)\\ n_{H_2SO_4\left(dư\right)}=\dfrac{10}{49}-0,02=\dfrac{451}{2450}\left(mol\right)\\ C\%H_2SO_4\left(dư\right)=\dfrac{\dfrac{451}{2450}.98}{101,6}.100=17,76\%\)

\(n_{Fe}=\dfrac{0.56}{56}=0.01\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(0.01.....0.02........0.01........0.01\)

\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)

\(V_{dd}=\dfrac{0.02}{1}=0.02\left(l\right)\)

\(C_{M_{FeCl_2}}=\dfrac{0.01}{0.02}=0.5\left(M\right)\)

Note : Đề sai hay thiếu chổ HCl ấy em.

\(n_{Fe}=\dfrac{0.56}{56}=0.01\left(mol\right)\)

\(n_{HCl}=0.1\cdot1=0.1\left(mol\right)\)

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

Ta có : 

\(\dfrac{n_{Fe}}{1}< \dfrac{n_{HCl}}{2}\Rightarrow HCldư\)

Khi đó : 

\(n_{Fe}=n_{FeCl_2}=n_{H_2}=0.01\left(mol\right)\)

\(V_{H_2}=0.01\cdot22.4=0.224\left(l\right)\)

\(b.\)

\(n_{HCl\left(dư\right)}=0.1-0.02=0.08\left(mol\right)\)

\(m_{HCl}=0.08\cdot36.5=2.92\left(\right)\)

\(c.\)

\(C_{M_{FeCl_2}}=\dfrac{0.01}{0.1}=0.1\left(M\right)\)

\(C_{M_{HCl\left(dư\right)}}=\dfrac{0.08}{0.1}=0.8\left(M\right)\)

\(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right);nHCl=0,1.1=0,1\left(mol\right)\\ PTHH:Fe+2HCl\rightarrow FeCl_2+H_2\\ Vì:\dfrac{0,01}{1}< \dfrac{0,1}{2}\\ \rightarrow HCldư\\ n_{H_2}=n_{Fe}=n_{FeCl_2}=0,01\left(mol\right)\\ V_{H_2\left(đktc\right)}=0,01.22,4=0,224\left(l\right)\\ V_{ddsau}=V_{ddHCl}=0,1\left(l\right)\\ C_{MddHCl}=\dfrac{0,01}{0,1}=0,1\left(M\right)\\ n_{HCl\left(dư\right)}=0,1-0,01.2=0,08\left(mol\right)\\ m_{HCl\left(dư\right)}=0,08.36,5=2,92\left(g\right)\\ C_{MddHCl\left(Dư\right)}=\dfrac{0,08}{0,1}=0,8\left(M\right)\)

1.

K2O +H2O -->2KOH(1)

a) muối trung hòa

2KOH +SO2 --> K2SO3+H2O(1)

n_K2O=0,1(mol)

theo(1) : nKOH=2nK2O=0,2(mol)

theo (2) : nSO2=1/2nKOH=0,1(mol)

=>mSO2=6,4(g)

b) muối axit :

KOH +SO2-->KHSO3(3)

theo(3) : nSO2=nKOH=0,2(mol)

=> mSO2=12,8(g)

c) cả 2 muối ...

ta có : nKHSO3=2nK2SO3

gọi nKHSO3=a(mol)

nK2SO3=b(mol)

=>\(\dfrac{a}{b}=\dfrac{2}{1}\)=>a-2b=0(I)

theo(2,3) : nKOH=a+2b=0,2(II)

từ(I,II) ta có :

\(\left\{{}\begin{matrix}a=0,1\left(mol\right)\\b=0,05\left(mol\right)\end{matrix}\right.\)

theo(2,3) : \(\Sigma nSO2\)=n_KHSO3,K2SO3=0,15(mol)

=>mSO2=9,6(g)

2.a) CaCl2 + 2AgNO3 --> Ca(NO3)2 + 2AgCl (1)

b ) nCaCL2=0,02(mol)

nAgNO3=0,01(mol)

lập tỉ lệ :

\(\dfrac{0,02}{1}>\dfrac{0,01}{2}\)

=>CaCl2 dư,AgNO3 hết => tính theo AgNO3

theo(1) : nAgCl=nAgNO3=0,01(mol)

=>mAgCl=1,435(g)

nCaCl2(pư)=nCa(NO3)2=1/2nAgNO3=0,005(mol)

=>mCa(NO3)2=0,82(g)

=>nCaCl2(dư)=0,015(mol)

=>mCaCl2(dư)=1,665(g)

(mik tính cả)

c) Vdd =30+70=100(ml)=0,1(l)

=> C_{Mdd CaCl2(dư)}=\(\dfrac{0,015}{0,1}=0,15\left(M\right)\)

C_{M dd Ca(NO3)2}=\(\dfrac{0,005}{0,1}=0,05\left(M\right)\)

C_{M AgCl}=\(\dfrac{0,01}{0,1}=0,1\left(M\right)\)

2)

a)......................CaCl₂ + 2 AgNO₃ ➞ 2 AgCl + Ca(NO₃)₂

Trước phản ứng:0.02..........0.01.............................................(mol)

Trong phản ứng:0.005.........0.01.............0.01...........0.005....(mol)

Sau phản ứng :...0.015............0...............0.01...........0.005....(mol)

b) n_CaCl2=2.22/111=0.02(mol)

n_AgNO3=1.7/170=0.01(mol)

m_{CaCl2 (dư)} = 0.015*111=1.665(g)

m_AgCl=0.01*143.5=1.435(g)

m_Ca(NO3)2=0.005*164=0.82(g)

c)V_dd=30+70=100(ml)

C_{M CaCl2} =0.015/0.1=0.15(M)

C_{M Ca(NO3)2}=0.005/0.1=0.05(M)

a) n_HNO3 = 0,2.0,25 = 0,05 mol

HNO₃ + KOH ---> KNO₃ + H₂O

0,05 0,05 0,05

V_KOH = 0,05/0,2 = 0,25 l

V_{dd KNO3} = 0,25 + 0,2 = 0,45 l

C_{M KNO3} = 0,05/0,45 = 0,(1) M

b) 2HNO₃ + Ba(OH)₂ ---> Ba(NO₃)₂ + 2H₂O

0,05 0,025 0,025

m_Ba(OH)2 = 0,025.171 = 4,275 g

m_{dd Ba(OH)2} = \(\frac{4,275.100\%}{20\%}\) = 21,375 g

m_Ba(NO3)2 = 0,025.261 = 6,525 g

C%_{dd Ba(NO3)2} = \(\frac{6,525}{21,375+200.1,02}.100\%\) \(\approx\) 2,9 %