Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

PT: \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

Theo PT: \(n_{H_2SO_4}=n_{MgSO_4}=n_{H_2}=m_{Mg}=0,2\left(mol\right)\)

a, \(m_{MgSO_4}=0,2.120=24\left(g\right)\)

b, \(V_{H_2}=0,2.24,79=4,958\left(l\right)\)

c, Bạn bổ sung đề phần này nhé.

\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ n_{MgSO_4}=n_{H_2}=n_{H_2SO_4}=n_{Mg}=0,2\left(mol\right)\\ a,m_{MgSO_4}=120.0,2=24\left(g\right)\\ b,V_{H_2\left(đkc\right)}=24,79.0,2=4,958\left(l\right)\\ c,Oxide:A_2O_x\left(x:hoá.trị.A\right)\\ A_2O_x+xH_2SO_4\rightarrow A_2\left(SO_4\right)_x+xH_2O\\ n_{Oxide}=\dfrac{\dfrac{3}{4}.0,2.1}{x}=\dfrac{0,15}{x}\left(mol\right)\\ M_{A_2O_x}=\dfrac{8}{\dfrac{0,15}{x}}=\dfrac{160}{3}x\)

Xét x=1;x=2;x=3;x=8/3 thấy x=3 (TM) khi đó KLR oxide là 160g/mol 

\(M_{M_2O_3}=2M_M+3.16=160\left(\dfrac{g}{mol}\right)\\ \Rightarrow M_M=\dfrac{160-48}{2}=56\left(\dfrac{g}{mol}\right)\)

Nên: M là sắt (Fe=56)

Oxide CTHH: Fe₂O₃

b) m_H2SO4 ban đầu là bao nhiêu :) ?

\(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

0,1        0,1             0,1             0,1 

\(a,m_{MgSO_4}=0,1.120=12\left(g\right)\)

\(b,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

\(c,m_{ddH_2SO_4}=\dfrac{0,1.98.100}{10}=98\left(g\right)\)

a, Magnesium + Sulfuric acid → Magnesium sulfate + Hydrogen

b, BTKL: m_Mg + m_H2SO4 = m_MgSO4 + m_H2

c, Từ b, có: m_H2SO4 = m_MgSO4 + m_H2 - m_Mg = 27,2 + 0,4 - 13 = 14,6 (g)

\(a.\) magnesium+sulfuric acid→magnesium sulfate+hydrogen

\(b.m_{Mg}+m_{H_2SO_4}=m_{MgSO_4}+H_2\\ c.\left(b\right)\Leftrightarrow13+m_{H_2SO_4}=27,2+0,4\\ \Leftrightarrow m_{H_2SO_4}=14,6g\)

đáp án a nhé 

a) Mg + H₂SO₄ --> MgSO₄ + H₂

b) \(n_{Mg}=\dfrac{14,4}{24}=0,6\left(mol\right)\)

PTHH: Mg + H₂SO₄ --> MgSO₄ + H₂

           0,6--->0,6------->0,6----->0,6

=> \(m_{H_2SO_4}=0,6.98=58,8\left(g\right)\)

c) 

PTHH: 2H₂ + O₂ --t^o--> 2H₂O

           0,6-->0,3

=> V_O2 = 0,3.24,79 = 7,437 (l)

=> V_kk = 7,437.5 = 37,185 (l)

nMg = 2,88/24 = 0,12 (mol)

PTHH: Mg + H2SO4 -> MgSO4 + H2

Mol: 0,12 ---> 0,12 ---> 0,12 ---> 0,12

mH2SO4 = 0,12 . 98 = 11,76 (g)

PTHH: 2H2 + O2 -> (t°) 2H2O

Mol: 0,12 ---> 0,06

Vkk = 0,06 . 5 . 24,79 = 7,437 (l)