\(n_{halogen}=\dfrac{35,875-18,625}{108-39}=0,25\left(mol\right)\\ M_{muối\left(t.gia\right)}=\dfrac{18,625}{0,25}=74,5=39+M_{halogen}\\ \Leftrightarrow M_{halogen}=35,5\left(\dfrac{g}{mol}\right)\\ Vậy.muối.halogen.đó:KCl\left(Cl=35,5\right)\\ Chọn.A\)

Bài 5 :

Độ tan của dd KNO₃ bão hòa là :

\(S=\dfrac{m_{ct}}{m_{dm}}.100\%=\dfrac{60}{190}.100\approx31,58\)

Bài 6 :

\(C\%_{NaCl}=\dfrac{5}{125}.100\%=4\%\)

1. \(C\%_{NaOH}=\dfrac{60}{300}.100\%=20\%\)

2. \(m_{HCl}=150.12\%=18\left(g\right)\)

3. \(m_{ddNa_2CO_3}=\dfrac{20}{15\%}=\dfrac{400}{3}\left(g\right)\)

1, \(2Na+2H_2O\rightarrow2NaOH+H_2\)

\(2KClO_3\underrightarrow{t^o}2KCl+3O_2\uparrow\)

2. Axit :

HCl : Axit clohidric 

HNO₃ : Axit Nitric

Bazo :

KOH : Kali hidroxit

Muối :

NaCl : Natri clorua 

BaCO₃ : Bari Cacbonat 

a, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)

b, \(n_{CH_4}=\dfrac{9,6}{16}=0,6\left(mol\right)\)

Theo PT: \(n_{CO_2}=n_{CH_4}=0,6\left(mol\right)\Rightarrow V_{CO_2}=0,6.22,4=13,44\left(l\right)\)

c, \(n_{O_2}=2n_{CH_4}=1,2\left(mol\right)\Rightarrow V_{O_2}=1,2.22,4=26,88\left(l\right)\)

\(\Rightarrow V_{kk}=\dfrac{V_{O_2}}{20\%}=134,4\left(l\right)\)

a, \(n_{NaOH}=0,2.0,3=0,06\left(mol\right)\)

\(m_{MgCl_2}=400.30\%=120\left(g\right)\Rightarrow n_{MgCl_2}=\dfrac{120}{95}=\dfrac{24}{19}\left(mol\right)\)

PT: \(2NaOH+MgCl_2\rightarrow Mg\left(OH\right)_2+2NaCl\)

Xét tỉ lệ: \(\dfrac{0,06}{2}< \dfrac{\dfrac{24}{19}}{1}\), ta được MgCl₂ dư.

Theo PT: \(n_{MgCl_2\left(pư\right)}=\dfrac{1}{2}n_{NaOH}=0,03\left(mol\right)\Rightarrow n_{MgCl_2\left(dư\right)}=\dfrac{24}{19}-0,03=\dfrac{2343}{1900}\left(mol\right)\)

\(\Rightarrow m_{MgCl_2\left(dư\right)}=\dfrac{2343}{1900}.95=117,15\left(g\right)\)

b, Theo PT: \(\left\{{}\begin{matrix}n_{NaCl}=n_{NaOH}=0,06\left(mol\right)\\n_{Mg\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=0,03\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}m_{NaCl}=0,06.58,5=3,51\left(g\right)\\m_{Mg\left(OH\right)_2}=0,03.58=1,74\left(g\right)\end{matrix}\right.\)

a, \(C_2H_6O+3O_2\underrightarrow{t^o}2CO_2+3H_2O\)

b, \(n_{CO_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{O_2}=\dfrac{3}{2}n_{CO_2}=0,3\left(mol\right)\)

\(\Rightarrow V_{O_2}=0,3.22,4=6,72\left(l\right)\)

c, \(n_{C_2H_6O}=\dfrac{1}{2}n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow m_{C_2H_6O}=0,1.46=4,6\left(g\right)\)

\(\Rightarrow V_{C_2H_6O}=\dfrac{4,6}{0,8}=5,75\left(ml\right)\)

Độ rượu = \(\dfrac{5,75}{50}.100=11,5^o\)

bạn viết mấy số mol dưới phương trình được không

\(n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

\(n_{C_2H_5OH}=\dfrac{9,2}{46}=0,2\left(mol\right)\)

PT: \(CH_3COOH+C_2H_5OH⇌CH_3COOC_2H_5+H_2O\) (xt, t^o)

Xét tỉ lệ: \(\dfrac{0,1}{1}< \dfrac{0,2}{1}\), ta được C₂H₅OH dư.

Theo PT: \(n_{CH_3COOC_2H_5}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOC_2H_5}=0,1.88=8,8\left(g\right)\)