PTHH: Mg+H₂SO₄-----> MgSO₄+H₂

\(n_{Mg}=\dfrac{4,8}{24}=0,2\) mol

\(n_{H_2SO_4}=\dfrac{14,7}{98}=0,15\) mol

Ta có tỉ lệ: \(\dfrac{0,2}{1}< \dfrac{0,15}{1}\)

----> Tính theo H₂SO₄

Theo pt: \(n_{H_2}=n_{H_2SO_4}=0,15\) mol

=> V_H2= 0,15.22,4= 3,36 l

a, PT: FeO+2HCl---->FeCl₂+H₂O

b, n_HCl=2.0,3=0,6 mol

Theo pt: n_FeO=\(\dfrac{1}{2}.n_{HCl}\)= \(\dfrac{1}{2}.0,6=0,3\) mol

=> m_FeO= 0,3.72= 21,6 g

c, theo pt: n_FeCl2=\(\dfrac{1}{2}.n_{HCl}=\dfrac{1}{2}.0,6=0,3\)mol

=> C_M dd FeCl₂=\(\dfrac{0,3}{0,3}=1\) M

a, PTHH: SO₂+H₂O----> H₂SO₃

CaO+H₂O----> Ca(OH)₂

Fe+2H₂O----> Fe(OH)₂+H₂

b, PTHH: CaO+H₂SO₄----> CaSO₄+H₂O

Fe+H₂SO₄---->FeSO₄+H₂

c, PTHH: SO₂+Ca(OH)₂----> CaSO₃+H₂O

Fe+Ca(OH)₂----> Fe(OH)₂+Ca

2HCl+Ca(OH)₂----> CaCl₂+2H₂O

a, \(x^4+4x^2-5\)

= \(x^4-x^2+5x^2-5\)

= \(x^2\left(x^2-1\right)+5\left(x^2-1\right)\)

= \(\left(x^2-1\right)\left(x^2+5\right)\)

= \(\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

b,\(x^3+5x^2+8x+4\)

= \(x^3+x^2+4x^2+4x+4x+4\)

= \(\left(x^3+x^2\right)+\left(4x^2+4x\right)+\left(4x+4\right)\)

= \(x^2\left(x+1\right)+4x\left(x+1\right)+4\left(x+1\right)\)

= \(\left(x+1\right)\left(x^2+4x+4\right)\)

= (\(x+1\))\(\left(x+2\right)^2\)

PTHH: 4P+5O₂---->2P₂O₅

n_P= 6,2/31 =0,2 (mol)

theo pt: n_P2O5= \(\dfrac{1}{2}\). n_P=\(\dfrac{1}{2}\).0,2=0,1 (mol)

=> m_P2O5= 0,1.142=14,2 (g)

FeBr₃

Fe₂(SO₄)₃

Fe₂(HPO₄)₃

Fe(NO₃)₃