a) \(3Fe+2O_2-t^o->Fe_3O_4\)

b) \(n_{Fe}=\frac{5,6}{56}=0,1\left(mol\right)\)

Theo pthh : \(n_{Fe_3O_4}=\frac{1}{3}n_{Fe_3O_4}=\frac{0,1}{3}\left(mol\right)\)

=> \(m_{Fe_3O_4}=232\cdot\frac{0,1}{3}\approx7,73\left(g\right)\)

c) Theo pthh : \(n_{O2\left(pứ\right)}=\frac{2}{3}n_{Fe}=\frac{0,2}{3}\left(mol\right)\)

=> \(n_{O2\left(can.dung\right)}=\frac{0,2}{3}\div100\cdot120=0,08\left(mol\right)\)

=> \(V_{O2\left(can.dung\right)}=0,08\cdot22,4=1,792\left(l\right)\)

1.\(4FeS_2+11O_2\rightarrow8SO_2+2Fe_2O_3\)

2.\(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

3. \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+NaSO_4\)

\(1.FeS_2+O_2---->SO_2+Fe_2O_3\)

 \(4FeS_2+11O_2\rightarrow8SO_2+2Fe_2O_3\)

\(2.Cu+HNO_3---->CU\left(NO_3\right)_2+NO+H_2O\)

  \(3Cu+8HNO_3\rightarrow3Cu\left(NO_3\right)_2+2NO+4H_2O\)

\(3.CuSO_4+NaOH---->Cu\left(OH\right)_2+Na_2SO_4\)

    \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

\(n_S=\frac{m}{M}=\frac{16}{32}=0,5\left(mol\right)\)

a,PTHH: \(S+O_2\rightarrow SO_2\)(có nhiệt độ nữa nhé)

(mol)    1         1          1

(mol)    0,5      0,5      0,5

b) Theo pt, ta có: \(n_S=n_{SO_2}=0,5\left(mol\right)\)

\(\Rightarrow m_{SO_2}=n.M=0,5.64=32\left(gam\right)\)

c)Theo pt, ta có: \(n_S=n_{O_2}=n_{SO_2}=0,5\left(mol\right)\)

\(\Rightarrow V_{O_2}\left(đktc\right)=n.22,4=0,5.22,4=11,2\left(lít\right)\)

\(\Rightarrow V_{SO_2}\left(đktc \right)=n.22,4=0,5.22,4=11,2\left(lit\right)\)

TL

a)\(S+O_2\underrightarrow{t^0}SO_2\)

b)\(n_S=\frac{16}{32}=0,5\left(mol\right)\)

\(S+O_2\underrightarrow{t^0}SO_2\)\(\)

0,5->0,5        0,5 (mol)

\(m_{SO_2}=0,5.64=32\left(g\right)\)

c)

\(V_{SO_2}=0,5.22,4=11,2\left(l\right)\)

jhhhhhhhhhhhhhhhhhhhhk

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Gọi x là hóa trị của R

Công thức dạng chung: R₂( SO₄)_x

%R= 28%

=>\(\dfrac{2R}{2R+96x}.100\%=28\%\)

=> \(\dfrac{R}{R+48x}.50\%=14\%\)

=> 50R= 14( R + 48x)

50R = 14R + 14.48x

=> 36R= 672x

=. R= \(\dfrac{672}{36}=\dfrac{56}{3}x\)

Nếu x=1=> R= \(\dfrac{56}{3}\)

       x=2 => R= \(\dfrac{112}{3}\)

       x=3 => R= 56

Vậy x =3 

R= 56( Fe )

CTHH: Fe₂( SO₄)_x

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Gọi 2 kim loại lần lượt là A,B hóa trị x, y

Ta thấy :

2A + xH₂SO₄ ------> A₂(SO₄)_x + xH₂

2B + yH₂SO₄ ------> B₂(SO₄)_y + yH₂

Ta có : n_H2SO4 = n_H2 = 2,8/22,4 = 0,125 ( mol)

=> m_H2SO4 = 0,125 . 98 = 12,25g

=> m_H2 = 0,125 . 2 = 0,25g

Áp dụng định luật bảo toàn khối lượng

m_KL= 18,8 + 0,25 - 12,25 = 6,8g

a. \(\cdot n_A=\frac{m_A}{M_A}\)

\(\cdot n=\frac{V}{22,4}\)

\(\cdot n=V.CM\)

\(\cdot n=\text{Số hạt vi mô}:6,023.10^{23}\)

\(\cdot V=n.22,4\)

\(\cdot CM=\frac{n}{V}\)

b. \(d_{A/B}=\frac{M_A}{M_B}\)

c. \(d_{A/KK}=\frac{M_A}{M_{KK}}=\frac{M_A}{29}\)