a)\(PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b)

\(n_{H2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PTHH:

\(n_{FeSO4}=n_{H2SO4}=n_{H2}=0,2\left(mol\right)\)

Ta có:

\(m_{FeSO4}=0,2.152=30,4\left(g\right)\)

\(m_{H2SO4}=0,2.98=19,6\left(g\right)\)

Fe + H₂SO₄ -> FeSO₄ + H₂

n_Fe=0,5(mol)

Theo PTHH ta có:

n_FeSO4=n_Fe=n_H2=n_H2SO4=0,5(mol)

m_FeSO4=152.0,5=76(g)

V_H2=22,4.0,5=11,2(lít)

Cách 1:

m_H2SO4=98.0,5=49(g)

Cách 2:

Theo ĐLBTKL ta có:

m_Fe + m_H2SO4=m_FeSO4+m_H2

=>m_H2SO4=76+0,5.2-28=49(g)

a.

n Fe==0,5 (mol)

Fe+H2SO4→FeSO4+H2↑

0,5→0,5         0,5      0,5        (mol)

b.

V H2(đktc)=0,5.22,4=11,2 (l)

c.

m HCl=0,5.36,5=18,25 (g) 

d.

m FeSO4=0,5.152=76 (g)

\(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\ n_{H_2SO_4}=n_{H_2}=n_{FeSO_4}=n_{Fe}=0,2\left(mol\right)\\ a,m_{H_2SO_4}=0,2.98=19,6\left(g\right)\\ b,C\%_{ddH_2SO_4}=\dfrac{19,6}{200}.100=9,8\%\\ c,m_{FeSO_4}=152.0,2=30,4\left(g\right)\\ d,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\)

a)

b)

nAl=23nH2=2,4(mol)⇒mAl=2,4.27=64,8(gam)

c)

Theo PT trên : 

a)

\(2Al + 3H_2SO_4 \to Al_2(SO_4)_3 + 3H_2\)

b)

\(n_{H_2} = \dfrac{6.13,44}{22,4} = 3,6(mol)\)

Theo PTHH : 

\(n_{Al} = \dfrac{2}{3}n_{H_2} = 2,4(mol)\\ \Rightarrow m_{Al} = 2,4.27 = 64,8(gam)\)

c)

\(4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\)

Theo PT trên : 

\(n_{O_2} = \dfrac{3}{4}n_{Al} = 1,8(mol)\\ \Rightarrow V_{O_2} = 1,8.22,4 = 40,32(lít)\)

Ta có: \(n_{H_2}=\dfrac{74,37}{24,79}=3\left(mol\right)\)

PT: \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

a, \(n_{Al}=\dfrac{2}{3}n_{H_2}=2\left(mol\right)\)

\(\Rightarrow m_{Al}=2.27=54\left(g\right)\)

b, \(n_{H_2SO_4}=n_{H_2}=3\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=3.98=294\left(g\right)\)

Gọi x, y lần lượt là số mol Al, Fe

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\\ Fe+H_2SO_4\rightarrow FeSO_4+H_2\\ n_{H_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\\ Tacó:\left\{{}\begin{matrix}27x+56y=0,83\\1,5x+y=0,025\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=0,01\\y=0,01\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}m_{Al}=0,27\left(g\right)\\m_{Fe}=0,56\left(g\right)\end{matrix}\right.\)

a) \(n_{H_2SO_4}=\frac{24,5}{98}=0,25\left(mol\right)\)

\(n_{H_2}=n_{H_2SO_4}=0,25\left(mol\right)\)

\(V_{H_2}=0,25.22,4=5,6l\)

b)m_H2=0,25.2=0.5g

nFe=0,4mol

Ta có: Fe dư, ax hết

nsp=nax=0,25mol

msp=0,25.152=38g

mFe dư: 22,4-(0,25.56)=8,4g

PTHH: Fe + H₂SO₄ -> FeSO₄ + H₂

Ta có:

\(n_{Fe}=\frac{22,4}{56}=0,4\left(mol\right)\\ n_{H_2SO_4}=\frac{24,5}{98}=0,25\left(mol\right)\)

Ta được:

\(\frac{0,4}{1}>\frac{0,25}{1}\)

=> H₂SO₄ hết, Fe dư nên tính theo \(n_{H_2SO_4}\)

a) Theo PTHH và đề bài, ta có:

\(n_{H_2}=n_{H_2SO_4}=0,25\left(mol\right)\)

Thể tích H₂ thu được (ở đktc):

\(V_{H_2\left(đktc\right)}=0,25.22,4=5,6\left(l\right)\)

b) Theo PTHH và đề bài, ta có:

\(n_{FeSO_4}=n_{H_2SO_4}=0,25\left(mol\right)\)

Khối lượng FeSO₄:

\(m_{FeSO_4}=n_{FeSO_4}.M_{FeSO_4}=0,25.152=38\left(g\right)\)

\(n_{Fe\left(phảnứng\right)}=n_{H_2SO_4}=0,25\left(mol\right)\)

=>\(n_{Fe\left(dư\right)}=n_{Fe\left(banđầu\right)}-n_{Fe\left(phảnứng\right)}=0,4-0,25=0,15\left(mol\right)\)

Khối lượng Fe dư:

\(m_{Fe\left(dư\right)}=n_{Fe\left(dư\right)}.M_{Fe}=0,15.56=8,4\left(g\right)\)

n_H2SO4=24,5/98=0,25(mol)

Fe + H₂SO₄ ------>FeSO₄ + H₂

TPT:n_H2=n_H2SO4=0,25(mol)

a)V_H2=0,25.22,4=5,6(L)

TPT:n_FeSO4=n_H2SO4=0,25(mol)

b)m_H2=0,25.2=0,5g

m_FeSO4=0,25.152=38g