a)  \(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)

PTHH: Zn + 2HCl --> ZnCl₂ + H₂

           0,4--------------------->0,4

=> V_H2 = 0,4.22,4 = 8,96 (l)

b) 

\(n_{Fe_2O_3}=\dfrac{16}{160}=0,1\left(mol\right)\)        

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,4}{3}\) => Fe₂O₃ hết, H₂ dư

PTHH: Fe₂O₃ + 3H₂ --t^o--> 2Fe + 3H₂O

                0,1---------------->0,2

=> m_Fe = 0,2.56 = 11,2 (g)

a, nZn = 26/65 = 0,4 (mol)

PTHH: Zn + 2HCl -> ZnCl2 + H2

nZn = nH2 = 0,4 (mol)

VH2 = 0,4 . 22,4 = 8,96 (l)

b, nFe2O3 = 16/160 = 0,1 (mol)

PTHH: Fe2O3 + 3H2 -> (t°) 2Fe + 3H2O

LTL: 0,1 < 0,4/3 => H2 dư

nFe = 0,1 . 3 = 0,3 (mol)

mFe = 0,3 . 56 = 16,8 (g)

\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

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

\(0.2.....................0.2..........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng }}=4.8+250-0.2\cdot2=254.4\left(g\right)\)

\(m_{MgCl_2}=0.2\cdot95=19\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{254.4}\cdot100\%=7.47\%\)

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

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____0,2____0,4______0,2____0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, Ta có: m _{dd sau pư} = m_Mg + m _{dd HCl} - m_H2 = 4,8 + 250 - 0,2.2 = 254,4 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{254,4}.100\%\approx7,47\%\)

Bạn tham khảo nhé!

Theo gt ta có: $n_{Mg}=0,2(mol)$

a, $Mg+2HCl\rightarrow MgCl_2+H_2$

b, Ta có: $n_{H_2}=0,2(mol)\Rightarrow V_{H_2}=4,48(l)$

c, Ta có: $m_{dd}=4,8+250-0,2.2=254,4(g)$

$\Rightarrow \%C_{MgCl_2}=7,47\%$

a, \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl → ZnCl₂ + H₂

Mol:     0,2                              0,2

(do Cu ko tác dụng với HCl loãng)

b, \(m_{Zn}=0,2.65=13\left(g\right)\)

  \(m_{Cu}=19,4-13=6,4\left(g\right)\)

Cu đứng sau H nên Cu không phản ứng được với dung dịch HCl

Fe + 2HCl → FeCl2 + H2

0,1 mol ← 0,1 mol

=> mFe = 0,1.56 = 5,6 gam => mCu = 12 – 5,6 = 6,4 gam

=>%mCu=6,4\12.100=53,3 %

=>%mFe=100-53,3=46,7%

a, PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

b, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

⇒ m_Zn = 0,2.65 = 13 (g)

⇒ m_Cu = 19,4 - 13 = 6,4 (g)

Bạn tham khảo nhé!

\(a) CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O\\ n_{CaCO_3} = n_{CO_2} = \dfrac{4,48}{22,4} = 0,2(mol)\\ m_{CaCO_3} = 0,2.100 = 20(gam)\\ b) n_{HCl} = 2n_{CO_2} = 0,4(mol)\\ \Rightarrow V_{dd\ HCl} = \dfrac{0,4}{3} = 0,13(lít)\)

CaCO₃ +  2HCl        → CaCl₂ + CO₂ + H₂O

   0.2    ←    0.4            ←           0.2

ⁿCO₂ = 4.48/22.4=0.2 mol

^mCaCO₃ = 0.2(40+12+16*3)=20g

v_HCl = 0.4/3=\(\dfrac{2}{15}\)lít

a, \(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

b, Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Theo PT: \(n_{Al}=\dfrac{2}{3}n_{H_2}=\dfrac{4}{15}\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{Al}=\dfrac{\dfrac{4}{15}.27}{10}.100\%=72\%\\\%m_{Cu}=28\%\end{matrix}\right.\)

c, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,4\left(mol\right)\)

\(\Rightarrow m_{H_2SO_4}=0,4.98=39,2\left(g\right)\)

\(\Rightarrow C\%_{H_2SO_4}=\dfrac{39,2}{300}.100\%\approx13,067\%\)