Bài 3: 

PTHH: \(2R+O_2\xrightarrow[]{t^O}2RO\)

Theo PTHH: \(n_R=n_{RO}\) \(\Rightarrow\dfrac{3,6}{M_R}=\dfrac{6}{M_R+16}\)

  \(\Rightarrow M_R=24\)  (Magie)

Bài 4 ở đâu vậy ??

bn ạ

phóng to màn hình lên là rõ ạ

\(2Al\left(OH\right)_3+3H_2SO_4\underrightarrow{trao..đổi}Al_2\left(SO4\right)_3+6H_2O\)

tỉ lệ số phân tử \(Al\left(OH\right)_3\):\(H_2SO_4\):\(Al_2\left(SO4\right)_3\):\(H_2O\) là 2:3:1:6

\(BTKL:mAL\left(OH\right)_3+mH_2SO_4=mAl_2\left(SO4\right)_3+mH_2O\)

                      23,4     \(+\)  \(mH_2SO_4\)  = 51,3             +  16,2

=>\(mH_2SO4=\left(51,3+16,2\right)-23,4=44,1\left(gam\right)\)                       

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bạn ơi bạn lấy từ đâu vậy ạ

a, \(n_{CaCO_3}=\dfrac{41,2}{100}=0,412\left(mol\right)\)

PTHH: CaO + H₂O → Ca(OH)₂

Mol:      0,412                 0,412                        

PTHH: Ca(OH)₂ + CO₂ → CaCO₃ + H₂O

Mol:       0,412        0,412          0,412

\(m_{CaO}=0,412.56=23,072\left(g\right)\)

b, \(V_{CO_2}=0,412.22,4=9,2288\left(l\right)\)

\(m_{Na_2CO_3}=100.16,96\%=16,96\left(g\right)\Rightarrow n_{Na_2CO_3}=\dfrac{16,96}{106}=0,16\left(mol\right)\)

\(m_{BaCl_2}=200.10,4\%=20,8\left(g\right)\Rightarrow n_{BaCl_2}=\dfrac{20,8}{208}=0,1\left(mol\right)\)

PTHH: Na₂CO₃ + BaCl₂ → BaCO₃ + 2NaCl

Mol:      0,1             0,1                           0,2

Ta có: \(\dfrac{0,16}{1}>\dfrac{0,1}{1}\) ⇒ Na₂CO₃ dư, BaCl₂ hết

m_{dd sau pứ} = 100 + 200 = 300 (g)

\(C\%_{ddNaCl}=\dfrac{0,1.58,5.100\%}{300}=1,95\%\)

\(C\%_{ddNa_2CO_3}=\dfrac{\left(0,16-0,1\right).106.100\%}{300}=2,12\%\)

....

1. a, PTK CO2: 12 + 16 . 2 = 44 (đvC)

b, PTK H2O: 2 + 16 = 18 (đvC)

c, PTK KNO3: 39 + 14 + 16 . 3 = 101 (đvC)

d, PTK Fe2(SO4)3: 56 . 2 + (32 + 16 . 4) . 3 = 400 (đvC)

e, PTK Br2: 80 . 2 = 160 (đvC)

2. a, CuO

b, Al2O3

c, Ba3(PO4)2

3. nFe = 5,6/56 = 0,1 (mol)

nCl2 = 10,65/71 = 0,15 (mol)

PTHH: 2Fe + 3Cl2 -> (t°) 2FeCl3

LTL: 0,1/2 = 0,15/3 => ko có chất dư

nFeCl3 = 0,1 (mol)

mFeCl3 = 0,1 . 162,5 = 16,25 (g)