Câu \(6\)

\(-\) acidic oxide:

\(+\)\(P_2O_5\):diphosphorus pentaoxide

\(-\) basic oxide:

\(+\) \(CuO\):copper \(\left(II\right)\) oxide 

\(-\) acid:

\(+\) \(HCl\):hydrogen chloride

\(+H_2SO_4\):sulfuric acid

\(+\)\(HNO_3\):nitrate acid

\(-\) base:

\(+\)\(Ca\left(OH\right)_2\):calcium hydroxide

\(+Fe\left(OH\right)_3\):iron \(\left(III\right)\) hydroxide

\(+Cu\left(OH\right)_2\):copper \(\left(II\right)\) hydroxide

\(-\) oxide trung tính:

\(+CO\):carbon monoxide

\(-\) muối:

\(+NaNO_3\):sodium nitrate

\(+BaSO_4\):barium sulfate

Câu \(7\)

\(PTHH:CaCO_3\xrightarrow[]{t^o}CaO+CO_2\)

Ta có: 40n_NaOH + 56n_KOH = 3,04 (1)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

\(KOH+HCl\rightarrow KCl+H_2O\)

Theo PT: \(n_{NaCl}=n_{NaOH}\)

\(n_{KCl}=n_{KOH}\)

⇒ 58,5n_NaOH + 74,5n_KOH = 4,15 (2)

Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{NaOH}=0,02\left(mol\right)\\n_{KOH}=0,04\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\%m_{KOH}=\dfrac{0,04.56}{3,04}.100\%\approx73,68\%\)

→ Đáp án: C

a

Chọn A

Ta có: \(n_{Mg}=\dfrac{9,6}{24}=0,4\left(mol\right)\)

PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,8\left(mol\right)\)

\(\Rightarrow C_{M_{HCl}}=\dfrac{0,8}{0,2}=4\left(M\right)\)

Ta có: \(m_{HCl}=200.7,3\%=14,6\left(g\right)\Rightarrow n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PT: \(K_2CO_3+2HCl\rightarrow2KCl+CO_2+H_2O\)

Theo PT: \(n_{CO_2}=n_{K_2CO_3}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)

a, V_CO2 = 0,2.24,79 = 4,958 (l)

b, \(C_{M_{K_2CO_3}}=\dfrac{0,2}{0,15}=\dfrac{4}{3}\left(M\right)\)

\(a.n_{Fe}=\dfrac{11,2}{56}=0,2mol\\ Fe+2HCl\rightarrow FeCl_2+H_2\)

0,2        0,4          0,2           0,2

\(V_{H_2}=0,2.24,79=4,958l\\ b.m_{HCl}=0,4.36,5=14,6g\\ c.m_{FeCl_2}=0,2.127=25,4g\)

bạn tách ra nhé

\(n_{Al}=\dfrac{31,6}{27}=\dfrac{158}{135}\left(mol\right)\)

PT : \(2Al+6H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

        \(\dfrac{158}{135}\)---->\(\dfrac{316}{45}\)--------->\(\dfrac{158}{135}\)---->\(\dfrac{158}{45}\)

a) \(m_{ddH2SO4}=\dfrac{\dfrac{316}{45}.98}{19,6\%}=3511,1\left(g\right)\)

b) \(V_{H2\left(đktc\right)}=\dfrac{158}{45}.22,4=78,65\left(l\right)\)

\(C\%_{Al2\left(SO4\right)3}=\dfrac{\dfrac{158}{135}.342}{31,6+3511,1-\dfrac{158}{45}.2}.100\%=11,32\%\)

anh cân bằng lại PTHH ạ

\(m_{ACl_3}=\dfrac{325.10}{100}=32,5g\\ n_{A_2O_3}=\dfrac{16}{2A+48}mol\\ n_{ACl_3}=\dfrac{32,5}{A+106,5}mol\\ A_2O_3+6HCl\rightarrow2ACl_3+3H_2O\\ \Rightarrow n_A=n_{ACl_3}:2\\ \Leftrightarrow\dfrac{16}{2A+48}=\dfrac{32,5}{A+106,5}:2\\ \Leftrightarrow A=56\)

Vậy A là Fe