mhh=\(\dfrac{3,36}{22,4}.64+\dfrac{2,8}{22,4}.28+\dfrac{6,72}{22,4}.2=13,7gam\)

=> ý A

a. 13,7 g

Ta có: \(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\) \(\Rightarrow m_{H_2}=0,35\cdot2=0,7\left(g\right)\)

Bảo toàn nguyên tố: \(n_{HCl}=2n_{H_2}=0,7\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,7\cdot36,5=25,55\left(g\right)\)

Bảo toàn khối lượng: \(m_{muối}=m_{KL}+m_{HCl}-m_{H_2}=34,55\left(g\right)\)

23a
24 ko có đấu ->

23.B

24.C

$23)$

$n_{Cl_2}=\dfrac{4,48}{22,4}=0,2(mol)$

$\Rightarrow m_{Cl_2}=0,2.71=14,2(g)$

$\to B$

$24)$

$Cu+2H_2SO_{4(đ)}\to CuSO_4+SO_4\uparrow+2H_2O$

Tỉ lệ: $1:2:1:1:2$

$\to C$

$25)CaCO_3\xrightarrow{t^o}CaO+CO_2\uparrow$

$\Rightarrow m_{CaCO_3}=m_{CO_2}+m_{CaO}$

$\Rightarrow m_{CaO}<m_{CaCO_3}\Rightarrow m_{rắn}$ giảm

$\to A$

$26)$ Bảo toàn KL:

$m_X=m_{oxit}+m_{CO_2}$

$\Rightarrow m_{oxit}=31,8-15,4=16,4(g)$

$\to B$

$27)$

$PTHH:4FeS_2+11O_2\xrightarrow{t^o}2Fe_2O_3+8SO_2\uparrow$

$\Rightarrow x:y=4:11$

$\to A$

$28)$

$n_{Fe}=\dfrac{140}{56}=2,5(mol)$

$Fe_2O_3+3CO\xrightarrow{t^o}2Fe+3CO_2\uparrow$

Theo PT: $n_{CO}=1,5.n_{Fe}=3,75(mol)$

$\Rightarrow V_{CO(đktc)}=3,75.22,4=84(lít)$

$\to $ Không đáp án nào đúng

\(1(đvC)=\dfrac{1}{12}.1,9926.10^{-23}=1,6605.10^{-24}(g)\\ \Rightarrow m_{Al}=27(đvC)=27.1,6605.10^{-24}\approx 4,48.10^{-23}(g)\)

Chọn C

6C

7A

8A

c

a

a

PTPU: Fe + O₂ -> Fe₂O₃

Áp dụng định luật BTKL:

m_Fe + m_O2 = m_Fe2O3

=> mFe2O3 = 23,2 - 16,8 = 6,4g

=> Chọn C

PTHH: \(3Fe+2O_2\overset{t^o}{--->}Fe_3O_4\)

Ta có: \(n_{Fe}=\dfrac{16,8}{56}=0,3\left(mol\right)\)

\(n_{Fe_3O_4}=\dfrac{23,2}{232}=0,1\left(mol\right)\)

Ta thấy: \(\dfrac{0,3}{3}=\dfrac{0,1}{1}\)

=> Không có chất dư.

Theo PT: \(n_{O_2}=2.n_{Fe_3O_4}=2.0,1=0,2\left(mol\right)\)

=> \(m_{O_2}=0,2.32=6,4\left(g\right)\)

Chọn C

\(\left\{{}\begin{matrix}n_{Cl_2}+n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\\\dfrac{n_{Cl_2}}{n_{O_2}}=\dfrac{1}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}n_{Cl_2}=0,1\left(mol\right)\\n_{O_2}=0,3\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{Cl_2}=0,1.71=7,1\left(g\right)\\m_{O_2}=0,3.32=9,6\left(g\right)\end{matrix}\right.\)

=> m_hh = 7,1 + 9,6 = 16,7(g)

Đặt $n_{Cl_2}=x(mol)\Rightarrow n_{O_2}=3x(mol)$

Mà $n_{hh}=n_{Cl_2}+n_{O_2}=\dfrac{8,96}{22,4}=0,4$

$\Rightarrow x+3x=0,4\Rightarrow x=0,1$

$\Rightarrow m_{Cl_2}=0,1.71=7,1(g);m_{O_2}=3.0,1.32=9,6(g)$

$\Rightarrow m_{hh}=7,1+9,6=16,7(g)$

TN1: (n_Mg;n_Al;n_Cu) = (a;b;c)

=> 24a + 27b + 64c = 14,2

PTHH: Mg + 2HCl --> MgCl₂ + H₂

______a---------------------------->a

2Al + 6HCl --> 2AlCl₃ + 3H₂

b-------------------------->1,5b

=> a + 1,5b = \(\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

TN2: (n_Mg;n_Al;n_Cu) = (2a;2b;2c)

PTHH: 2Mg + O₂ --t^o--> 2MgO

______2a--->a

4Al + 3O₂ --t^o--> 2Al₂O₃

2b--->1,5b

2Cu + O₂ --t^o--> 2CuO

2c--->c

=> a + 1,5b + c = \(\dfrac{11,2}{22,4}=0,5\)

=> a=0,1 (mol); b = 0,2 (mol); c = 0,1(mol
=> \(\%Mg=\dfrac{0,1.24}{14,2}.100\%=16,9\%\)