PTHH : 2Al     +     6HCl  --> 2AlCl3   +    3H2 ↑   (1)

n_AlCl3 = \(\dfrac{m}{M}=\dfrac{13,35}{27+35,5.3}=0.1\left(mol\right)\) 

Từ (1) => n_HCl   =   2n_H2  = 0.2 (mol)

=> m_HCl = n.M  =  0.2 x  36.5 = 7.3 (g)

\(PTHH:2Al+6HCl\rightarrow2AlCl_3+3H_2\\ n_{AlCl_3}=\dfrac{m}{M}=\dfrac{13,35}{133,5}=0,1\left(mol\right)\\ Theo.PTHH:n_{HCl}=3.n_{AlCl_3}=3.0,1=0,3\left(mol\right)\\ m_{HCl}=n.M=0,3.36,5=10,95\left(g\right)\)

\(n_{O_2}=\dfrac{3.36}{22.4}=0.15\left(mol\right)\)

\(3Fe+2O_2\underrightarrow{t^0}Fe_3O_4\)

\(.......0.13....0.075\)

\(V_{O_2}=3.36\left(l\right)\)

\(m_{Fe_3O_4}=0.075\cdot232=17.4\left(g\right)\)

a) pt: 3Fe + 2O₂ \(\rightarrow\) Fe₃O₄

b)  Thể tích khí oxi cho ở đề bài rồi mà

c) Theo pt: nFe₃O₄ = \(\dfrac{1}{2}n_{O_2}=\dfrac{1}{2}.0,15=0,075mol\)

\(\Rightarrow mFe_3O_4=0,075.232=17,4g\)

Em chụp lại c8

đây ạ

Fe3O4 + 3H2 -> 3Fe + 4H2O

  232                  3\(\times\)56              (M)

\(mFe3O4=1.5\times80\%=1.2\) tấn

\(mFe=\dfrac{1.2\times3\times56}{232}=0.87\) tấn

Chọn D

Câu 5 : 

a)

$n_{NaOH} = 0,1.2,5 = 0,25(mol)$
$m_{NaOH} = 0,25.40 = 10(gam)$

b)

$C_{M_{CaCl_2}} = \dfrac{0,02}{0,2} = 0,1M$
c)

$C\%_{NaCl} = \dfrac{6}{6 + 144}.100\% = 4\%$

Câu 6 : 

a) $2Na + 2H_2O \to 2NaOH + H_2$

$n_{H_2} = \dfrac{3,36}{22,4} = 0,15(mol)$
$n_{NaOH} = 2n_{H_2} = 0,3(mol)$
$m_{NaOH} = 0,3.40 = 12(gam)$

b)

$CuO + H_2 \xrightarrow{t^o} Cu + H_2O$

$n_{Cu} = n_{H_2} = 0,15(mol)$
$m_{Cu} = 0,15.64 = 9,6(gam)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

            \(FeO+H_2\xrightarrow[]{t^o}Fe+H_2O\)

Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)=n_{H_2}=n_{Fe}\) \(\Rightarrow m_{Fe}=0,1\cdot56=5,6\left(g\right)\)

Bạn cần giải câu nào?

\(a.\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(b.\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(a.........................a\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(\dfrac{2}{3}a........................a\)

\(m_{Mg}=24a\left(g\right)\)

\(m_{Al}=\dfrac{2}{3}a\cdot27=18a\left(g\right)\)

\(m_{Al}< m_{Mg}\)

=> Cần dùng Al là nhỏ nhất

\(2.\)

\(CuO+H_2\underrightarrow{^{t^0}}Cu+H_2O\)

\(Fe_2O_3+3H_2\underrightarrow{^{t^0}}2Fe+3H_2O\)

\(ZnO+H_2\underrightarrow{^{t^0}}Zn+H_2O\)

\(n_{H_2O}=n_{H_2}=\dfrac{5.4}{18}=0.3\left(mol\right)\)

\(BTKL:\)

\(m_A+m_{H_2}=m_{kl}+m_{H_2O}\)

\(\Rightarrow m_{kl}=20+0.3\cdot2-5.4=15.2\left(g\right)\)