\(n_{Na}=\dfrac{6.9}{23}=0.3\left(mol\right)\)

\(n_S=\dfrac{3.84}{32}=0.12\left(mol\right)\)

\(2Na+S\underrightarrow{^{^{t^0}}}Na_2S\)

Lập tỉ lệ : \(\dfrac{0.3}{2}>\dfrac{0.12}{1}\Rightarrow Nadư\)

A gồm : Na₂S , Na

\(m_{Na\left(dư\right)}=\left(0.3-0.12\cdot2\right)\cdot23=1.38\left(g\right)\)

\(m_{Na_2S}=0.12\cdot78=9.36\left(g\right)\)

\(b.\)

\(Na+HCl\rightarrow NaCl+\dfrac{1}{2}H_2\)

\(Na_2S+2HCl\rightarrow2NaCl+H_2S\)

\(V=0.03\cdot22.4+0.12\cdot22.4=3.36\left(l\right)\)

a)

$n_{Na} = \dfrac{6,9}{23} = 0,3(mol)$
$n_S = \dfrac{3,84}{32} = 0,12(mol)$
$2Na + S \xrightarrow{t^o} Na_2S$

$n_{Na} : 2 = 0,15 > n_S : 1$ do đó Na dư.

$n_{Na_2S} =n_S = 0,12(mol)$
$n_{Na\ dư} = 0,3 - 0,12.2 = 0,06(mol)$
$m_{Na_2S} = 0,12.78 = 9,36(gam)$
$m_{Na\ dư} = 0,06.23 = 1,38(gam)$

b)
$2Na + 2HCl \to 2NaCl + H_2$
$Na_2S + 2HCl \to 2NaCl + H_2S$
$n_{H_2} = \dfrac{1}{2}n_{Na\ dư} =  0,03(mol)$
$n_{H_2S} = n_{Na_2S} = 0,12(mol)$
$V_{H_2} = (0,03 + 0,12).22,4 = 3,36(lít)$

Đáp án D

a, PT: \(Fe+S\underrightarrow{t^o}FeS\)

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

\(n_S=\dfrac{3,2}{32}=0,1\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,3}{1}>\dfrac{0,1}{1}\), ta được Fe dư.

Chất rắn A gồm Fe dư và FeS.

Theo PT: \(n_{Fe\left(pư\right)}=n_{FeS}=n_S=0,1\left(mol\right)\)

\(\Rightarrow n_{Fe\left(dư\right)}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(FeS+2HCl\rightarrow FeCl_2+H_2S\)

Theo PT: \(\left\{{}\begin{matrix}n_{H_2}=n_{Fe\left(dư\right)}=0,2\left(mol\right)\\n_{H_2S}=n_{FeS}=0,1\left(mol\right)\end{matrix}\right.\)

Ở cùng điều kiện nhiệt độ và áp suất, %V cũng là % số mol.

\(\Rightarrow\left\{{}\begin{matrix}\%V_{H_2}=\dfrac{0,2}{0,2+0,1}.100\%\approx66,67\%\\\%V_{H_2S}\approx33,33\%\end{matrix}\right.\)

b, Ta có: \(\Sigma n_{HCl\left(dadung\right)}=2n_{Fe}+2n_{FeS}=0,6\left(mol\right)\) (1)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có: \(n_{NaOH}=0,1.2=0,2\left(mol\right)\)

\(\Rightarrow n_{HCl\left(dư\right)}=n_{NaOH}=0,2\left(mol\right)\) (2)

Từ (1) và (2) \(\Rightarrow\Sigma n_{HCl}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{ddHCl}}=\dfrac{0,8}{0,5}=1,6\left(M\right)\)

Bạn tham khảo nhé!

nCO2 =nCa(OH)2=6/100=0,06 mol=nCO pứ
nFe2O3=16/160=0,1 mol
=>nFe=0,2 mol
bảo toàn Fe nFe hh sau pứ=0,2 mol
bảo toàn klg=> m cr sau pứ=16+0,06.28-0,06.44=15,04 gam
GS hh cr sau pứ gồm Fe và O
=>mO=15,04-0,2.56=3,84 gam
=>nO=0,24 mol
khi cho hh cr tác dụng với H2SO4 đặc nóng
O          +2e     => O−2
0,24 mol=>0,48 mol
S+6 +2e     => S+4
          0,12 mol=>0,06 mol
Fe             => Fe+3 +3e
0,2 mol                      =>0,6 mol
VSO2=0,06.22,4=1,344 lit

c giải cho e r mà?

nCO2=nCa(OH)2=6/100=0,06 mol=nCO

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C1 : 

- Hòa tan hh vào dd HCl : 

Mg + 2HCl => MgCl2 + H2 

Fe + 2HCl => FeCl2 + H2 

X : MgCl2 , FeCl2 , HCl dư 

Y : Cu 

Z : H2 

- Dung dịch X + NaOH : 

MgCl2 + 2NaOH => Mg(OH)2 + 2NaCl

FeCl2 + 2NaOH => Fe(OH)2 + 2NaCl 

Kết tủa T : Mg(OH)2 , Fe(OH)2 

- Nung T : 

Mg(OH)2 -to-> MgO + H2O 

4Fe(OH)2 + O2 -to-> 2Fe2O3 + 4H2O

Chất rắn : MgO , Fe2O3

C2:

Đặt : nCl2 = x (mol) , nO2 = y (mol) 

nA = x + y = 0.6 (mol) (1) 

m_Cl2 + m_O2 = 48.15 - 19.2 = 28.95 (g) 

=> 71x + 32y = 28.95 (2) 

(1),(2) : 

x = 0.25 , y = 0.35 

Đặt : nMg = a (mol) , nAl = b (mol) 

Mg => Mg+2 + 2e 

Al => Al+3 + 3e 

Cl2 + 2e => 2Cl-1 

O2 + 4e => 2O2- 

BT e : 

2a + 3b = 0.25*2 + 0.35*4 = 1.9 

mB = 24a + 27b = 19.2 

=> a = 0.35 

b = 0.4 

%Mg = 0.35*24/19.2 * 100% = 43.75%

ko có ai giúp tôi à

                         n_Fe = m/M= 16,8/56=0,3(mol)

                         n_S = m/M= 6,4/32=0,2(mol)

                PTHH:   Fe   +  S  =   FeS

Trước phản ứng:  0,3      0,3      0,3      (mol)

          Phản ứng:  0,3      0,2                  (mol)

   Sau phản ứng:   0        0,1      0,3       (mol)

a/ chất rấn A gồm Fe và S

b/            PTHH:  FeS + 2HCl = Fecl₂ + H₂S

Theo phản ứng:    1          2          1           1      (mol)

      Theo bài ra:    0,3      0,6       0,3        0,3    (mol)

v_H2S = n.22,4=0,3.22,4=6,72(l)

đun nóng 16.8 g bột sắt và 6.4 g bột lưu huỳnh (không có không khí) thu được chất rắn A

a)chất rắn A gồm những chất gì (biết Fe+S=FeS) tính khới lượng từng chất trong A

b)cho chất rắn A tác dụng với HCl dư thì còn những  phản ứng hóa học sảy ra thính thể tích khí thoát ra ở điều kiệ

Theo đề bài ta có : nkt = nBaCO3 = 1,97/197 = 0,01 (mol)

PTHH :

FeO+CO−t0−>Fe+CO2↑FeO+CO−t0−>Fe+CO2↑

Fe2O3+3CO−t0−>2Fe+3CO2↑Fe2O3+3CO−t0−>2Fe+3CO2↑

Fe3O4+4CO−t0−>3Fe+4CO2↑Fe3O4+4CO−t0−>3Fe+4CO2↑

CO2 + Ba(OH)2 - > BaCO3 + H2O

0,01mol........................0,01mol

Theo các PTHH ta có : nCo = nCo2 = 0,01 mol

Áp dụng đlbtkl ta có :

mX + mCO = mY + mCO2

=> mY = 4,64 + 0,01.28 - 0,01.44 = 4,48(g)

Vậy....

Đề hỏi gì á em?

Theo ĐLBT KL: m _X + m_O2 = m_Y

⇒ m_O2 = 8,7 - 6,7 = 2 (g)

\(\Rightarrow n_{O_2}=\dfrac{2}{32}=0,0625\left(mol\right)\) \(\Rightarrow V_{O_2}=0,0625.22,4=1,4\left(l\right)\)

\(\Rightarrow V_{kk}=5V_{O_2}=7\left(l\right)\)