Hóa học

\(m_{NaCl}=50.60\%=30\left(g\right)\)

\(m_{H_2O\left(thêm\right)}=150.1=150\left(g\right)\)

=> m_{dd(sau khi thêm)} = 150 + 50 = 200 (g)

=> \(C\%_{dd.sau.khi.thêm}=\dfrac{30}{200}.100\%=15\%\)

a) \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

b) Gọi số mol Al₂O₃, CuO là a, b (mol)

=> 102a + 80b = 18,2 (1)

PTHH: \(CuO+CO\underrightarrow{t^o}Cu+CO_2\)

              b------------->b---->b

=> 102a + 64b = 16,6 (2)

(1)(2) => a= 0,1 (mol); b = 0,1 (mol)

=> V = 0,1.22,4 = 2,24 (l)

c) \(n_{Ba\left(OH\right)_2}=1.0,07=0,07\left(mol\right)\) 

PTHH: \(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

                0,07----->0,07----->0,07

             \(BaCO_3+CO_2+H_2O\rightarrow Ba\left(HCO_3\right)_2\)

                 0,03<---0,03

=> \(m_{BaCO_3}=\left(0,07-0,03\right).197=7,88\left(g\right)\)

a) \(CuO+CO\xrightarrow[]{t^o}Cu+CO_2\)

b) \(m_{giảm}=m_O=18,2-16,6=1,6\left(g\right)\)

\(n_{CO_2}=n_O=\dfrac{1,6}{16}=0,1\left(mol\right) \)

=> V = 0,1.22,4 = 2,24 (l)

c) \(n_{Ba\left(OH\right)_2}=1.0,07=0,07\left(mol\right)\)

Xét \(T=\dfrac{n_{Ba\left(OH\right)_2}}{n_{CO_2}}=\dfrac{0,07}{0,1}=0,7\) => Tạo cả 2 muối

\(Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\)

0,07--------->0,07------>0,07

\(BaCO_3+CO_2+H_2O\rightarrow Ba\left(HCO_3\right)_2\)

0,03<-----0,03

=> m_kt = (0,07 - 0,03).197 = 7,88 (g)

\(a.CuO+CO-^{t^o}\rightarrow Cu+CO_2\\ b.m_{giảm}=m_{O\left(trongCuO\right)}=18,2-16,6=1,6\left(g\right)\\ \Rightarrow n_{CO_2}=n_O=\dfrac{1,6}{16}=0,1\left(mol\right)\\ \Rightarrow V_{CO_2}=0,1.22,4=2,24\left(l\right)\\ c.n_{Ba\left(OH\right)_2}=0,07\left(mol\right)\\ LTL:\dfrac{n_{CO_2}}{n_{Ba\left(OH\right)_2}}=\dfrac{0,1}{0,07}=1,42\Rightarrow Xảyra2phảnứng\\ PTHH:Ba\left(OH\right)_2+CO_2\rightarrow BaCO_3+H_2O\\ Ba\left(OH\right)_2+2CO_2\rightarrow Ba\left(HCO_3\right)_2\\Đặt:\left\{{}\begin{matrix}n_{BaCO_3}=x\left(mol\right)\\n_{Ba\left(HCO_3\right)_2}=y\left(mol\right)\end{matrix}\right.\\ Tacó:\left\{{}\begin{matrix}x+y=0,07\\x+2y=0,1\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,04\\y=0,03\end{matrix}\right.\\ \Rightarrow m_{BaCO_3}=0,04.197=7,88\left(g\right)\)

a) \(Fe_xO_y+yCO\xrightarrow[]{t^o}xFe+yCO_2\)

b) \(n_O=n_{CO_2}=\dfrac{3,136}{22,4}=0,14\left(mol\right)\)

\(BTNT:m_{Fe}=8,12-0,14.16=5,88\left(g\right)\\ d,n_{Fe}=\dfrac{5,88}{56}=0,105\left(mol\right)\\ \Rightarrow\dfrac{x}{y}=\dfrac{n_{Fe}}{n_O}=\dfrac{0,105}{0,14}=\dfrac{3}{4}\)

=> CTHH: `Fe_3O_4` sắt từ oxit

a) 

b) BTNT:mFe=8,12−0,14.16=5,88(g)d,nFe=5,8856=0,105(mol)⇒xy=nFenO=0,1050,14=34BTNT:mFe=8,12−0,14.16=5,88(g)d,nFe=5,8856=0,105(mol)⇒xy=nFenO=0,1050,14=34

=> CTHH:  sắt từ oxit

\(\%O_{\left(SO_2\right)}=\dfrac{32}{64}.100\%=50\%\\ \%O_{\left(SO_3\right)}=\dfrac{48}{80}.100\%=60\%\\ \%O_{\left(CO_2\right)}=\dfrac{32}{44}.100\%=72,73\%\\ \%O_{\left(NO_2\right)}=\dfrac{32}{46}.100\%=69,57\%\)

=> A

\(n_{NaOH}=0,2.2=0,4\left(mol\right)\)

`a)`\(2NaOH+CuCl_2\rightarrow2NaCl+Cu\left(OH\right)_2\downarrow\)

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

`b)`\(m_{NaCl}=0,4.58,5=23,4\left(g\right)\)

`c)`\(m_{Cu\left(OH\right)_2}=0,2.98=19,6\left(g\right)\)

`d)`\(m_{dd_{CuCl_2}}=\dfrac{0,2.135}{20\%}=135\left(g\right)\)

a) `n_{NaOH} = 0,2.2 = 0,4 (mol)`

PTHH: `2NaOH + CuCl_2 -> Cu(OH)_2 + 2NaCl`

              0,4------>0,2--------->0,2--------->0,4

b) `m_{NaCl} = 0,4.58,5 = 23,4 (g)`

c) `m_{Cu(OH)_2} = 0,2.98 = 19,6 (g)`

d) `m_{ddCuCl_2} = {0,2.135}{20%} = 135 (g)`

CTHH của oxit là: `R_2O_3`

Ta có: \(\dfrac{16.3}{16.3+2M_R}.100\%=30\%\)

=> M_R = 56 => R là Fe

CTHH: `Fe_2O_3`

Ta có: \(\%O=\dfrac{16x}{32+16x}.100\%=60\%\)

=> x = 3

=> CTHH: SO₃

200 gam dung dịch $H_2SO_4$ 5M có : 

$V_{dd\ H_2SO_4} = 200 : 1,29 = 155(ml)$

Sau khi trộn : 

$n_{H_2SO_4} = 0,15.2 + 0,155.5 = 1,075(mol)$
$V_{dd} = 0,15 + 0,155 = 0,305(lít)$
$C_{M_{H_2SO_4}} = \dfrac{1,075}{0,305} = 3,52M$

- Gọi số mol NaHCO3 và Na2CO3 là x,y

PT : CO2  + NaOH ----> NaHCO3 (1)

         x           x                  x

       CO2 + 2NaOH ----> Na2CO3 + H2O (2)

         y           2y                 y

- n_{CO_2}=\frac{15,68}{22,4}=0,7\left(n\right)nCO2​​=22,415,68​=0,7(n)

- Ta có : x + y = 0,7 ( mol ) => y = 0,7 - x (1)

               84x + 106y = 65,4 (2)

- Thay (1) vào (2) , ta có : 

      84x + 106(0,7 - x ) = 65,4

<=> 84x + 74,2 - 106x = 65,4

<=> -22x = 65,4 - 74,2

<=> -22x = -8,8

<=> 22x = 8,8

<=>    x  = 0,4 => y = 0,3

- n_{NaOH}=0,4+2\times0,3=1\left(n\right)nNaOH​=0,4+2×0,3=1(n)

\Rightarrow C_{M_{NaOH}}=\frac{1}{0,5}=2\left(M\right)⇒AMNaOH​​=0,51​=2(M)

Nếu sai ko chửi tui nha

\(n_{CO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\); \(m_{dd.NaOH}=164.1,22=200\left(g\right)\)

\(n_{NaOH}=\dfrac{200.20\%}{40}=1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{n_{NaOH}}{n_{CO_2}}=\dfrac{1}{0,25}=4\) => Tạo muối Na₂CO₃

PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

                 0,5<----0,25------>0,25

=> \(\left\{{}\begin{matrix}m_{Na_2CO_3}=0,25.106=26,5\left(g\right)\\m_{NaOH\left(dư\right)}=\left(1-0,5\right).40=20\left(g\right)\end{matrix}\right.\)