Ta có: \(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

PT: \(CuO+2HCl\rightarrow CuCl_2+H_2O\)

a, Theo PT: \(n_{CuCl_2}=n_{CuO}=0,1\left(mol\right)\Rightarrow m_{CuCl_2}=0,1.135=13,5\left(g\right)\)

b, \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\Rightarrow C_{M_{HCl}}=\dfrac{0,2}{0,2}=1\left(M\right)\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, Ta có: \(n_{H_2SO_4}=0,5.1=0,5\left(mol\right)\)

Theo PT: \(n_{Fe}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow m_{Fe}=0,5.56=28\left(g\right)\)

c, \(n_{FeSO_4}=n_{H_2SO_4}=0,5\left(mol\right)\Rightarrow C_{M_{FeSO_4}}=\dfrac{0,5}{0,5}=1\left(M\right)\)

CTHH sai và sửa lại:

HO → H₂O

CaOH → Ca(OH)₂

Fe₂O → FeO hoặc Fe₂O₃ hoặc Fe₃O₄

NaO → Na₂O

K₂NO₃ → KNO₃

Ca₂(PO₄)₃ → Ca₃(PO₄)₂

Phần giải thích bạn xem lại quy tắc hóa trị để tự giải thích nhé.

HO sai vì H hoá trị l O hoá trị ll

⇒CTHH: H₂O

CaOH sai vì Ca hoá trị ll OH hoá trị l

⇒CTHH: Ca(OH)₂

Fe₂O sai vì sai vì Fe hoá trị ll,lll O hoá trị ll

⇒CTHH: FeO; Fe₂O₃

NaO sai vì Na hoá trị l O hoá trị ll

⇒CTH:Na₂O

K₂NO₃ sai vì K hoá trị l NO₃ hoá trị l

⇒CTHH: KNO₃

Ta có: \(m_{H_2SO_4}=50.9,8\%=4,9\left(g\right)\Rightarrow n_{H_2SO_4}=\dfrac{4,9}{98}=0,05\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

a, \(n_{Fe}=n_{H_2SO_4}=0,05\left(mol\right)\Rightarrow m_{Fe}=0,05.56=2,8\left(g\right)\)

b, \(n_{FeSO_4}=n_{H_2}=n_{H_2SO_4}=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,8 + 50 - 0,05.2 = 52,7 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{0,05.152}{52,7}.100\%\approx14,42\%\)

CTHH sai và sửa lại:

K₂ → K

Al₃O₂ → Al₂O₃

O → O₂

Ca_zC₁ → CaC₂

(OH)Fe → Fe(OH)₂ hoặc Fe(OH)₃

MgNO → Mg(NO₃)₂

K(OH) → KOH

P_sO_z → P₂O₅

H → H₂

MgNO→Mg(NO₃)₂ chứ cj Mg hoá trị ll mà

Ta có: M_K2CO3 = 39.2 + 12 + 16.3 = 138 (g/mol)

\(\Rightarrow\left\{{}\begin{matrix}\%K=\dfrac{39.2}{138}.100\%\approx56,5\%\\\%C=\dfrac{12}{138}.100\%\approx8,7\%\\\%O\approx34,8\%\end{matrix}\right.\)

a, Ta có: \(n_{Fe}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

Theo PT: \(n_{FeSO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow m_{FeSO_4}=0,1.152=15,2\left(g\right)\)

b, \(n_{H_2}=n_{Fe}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)

c, Sửa đề: 500 ml → 500 (g)

Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,1\left(mol\right)\Rightarrow C\%_{H_2SO_4}=\dfrac{0,1.98}{500}.100\%=1,96\%\)

Ta có: \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

PT: \(Fe+2HCl\rightarrow FeCl_2+H_2\)

a, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)

b, \(n_{HCl}=2n_{Fe}=0,4\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

c, \(n_{FeCl_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{FeCl_2}=0,2.127=25,4\left(g\right)\)

a, \(Fe+2HCl\rightarrow FeCl_2+H_2\)

\(n_{Fe}=\dfrac{2,8}{56}=0,05\left(mol\right)\)

\(m_{HCl}=43,8.10\%=4,38\left(g\right)\Rightarrow n_{HCl}=\dfrac{4,38}{36,5}=0,12\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,05}{1}< \dfrac{0,12}{2}\), ta được HCl dư.

Theo PT: \(n_{H_2}=n_{Fe}=0,05\left(mol\right)\Rightarrow V_{H_2}=0,05.22,4=1,12\left(l\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{FeCl_2}=n_{Fe}=0,05\left(mol\right)\\n_{HCl\left(pư\right)}=2n_{Fe}=0,1\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow n_{HCl\left(dư\right)}=0,12-0,1=0,02\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,8 + 43,8 - 0,05.2 = 46,5 (g)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{FeCl_2}=\dfrac{0,05.127}{46,5}.100\%\approx13,66\%\\C\%_{HCl\left(dư\right)}=\dfrac{0,02.36,5}{46,5}.100\%\approx1,57\%\end{matrix}\right.\)

a, \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)

b, \(n_{Fe}=\dfrac{11,2}{56}=0,2\left(mol\right)\)

Theo PT: \(n_{H_2SO_4}=n_{Fe}=0,2\left(mol\right)\Rightarrow m_{H_2SO_4}=0,2.98=19,6\left(g\right)\)

c, \(C\%_{H_2SO_4}=\dfrac{19,6}{50}.100\%=39,2\%\)

d, Theo PT: \(n_{H_2}=n_{Fe}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.22,4=4,48\left(l\right)\)