a, \(n_{CO_2}=\dfrac{0,56}{22,4}=0,025\left(mol\right)\)

PT: \(2CH_3COOH+Na_2CO_3\rightarrow2CH_3COONa+CO_2+H_2O\)

Theo PT: \(n_{CH_3COOH}=2n_{CO_2}=0,05\left(mol\right)\)

\(\Rightarrow C\%_{CH_3COOH}=\dfrac{0,05.60}{100}.100\%=3\%\)

b, Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,025\left(mol\right)\Rightarrow m_{Na_2CO_3}=0,025.106=2,65\left(g\right)\)

\(n_{CH_3COONa}=2n_{CO_2}=0,05\left(mol\right)\Rightarrow m_{CH_3COONa}=0,05.82=4,1\left(g\right)\)

c, \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,05\left(mol\right)\)

Mà: H = 80%

\(\Rightarrow n_{C_2H_5OH\left(TT\right)}=\dfrac{0,05}{80\%}=0,0625\left(mol\right)\)

\(\Rightarrow m_{C_2H_5OH\left(TT\right)}=0,0625.46=2,875\left(g\right)\)

\(\Rightarrow V_{C_2H_5OH}=\dfrac{2,875}{0,8}=3,59375\left(ml\right)\)

\(\Rightarrow V_{C_2H_5OH\left(10^o\right)}=\dfrac{3,59375}{10}.100=35,9375\left(ml\right)\)

a, \(m_{CH_3COOH}=100.6\%=6\left(g\right)\Rightarrow n_{CH_3COOH}=\dfrac{6}{60}=0,1\left(mol\right)\)

PT: \(CH_3COOH+NaHCO_3\rightarrow CH_3COONa+CO_2+H_2O\)

Theo PT: \(n_{NaHCO_3}=n_{CH_3COONa}=n_{CO_2}=n_{CH_3COOH}=0,1\left(mol\right)\)

\(\Rightarrow m_{NaHCO_3}=0,1.84=8,4\left(g\right)\)

\(V_{CO_2}=0,1.22,4=2,24\left(l\right)\)

b, Ta có: m _{dd sau pư} = 100 + 8,4 - 0,1.44 = 104 (g)

\(\Rightarrow C\%_{CH_3COONa}=\dfrac{0,1.82}{104}.100\%\approx7,88\%\)

a, \(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\)

PT: \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

Theo PT: \(n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\)

\(\Rightarrow a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\)

b, \(m_{dd}=\dfrac{15}{2\%}=750\left(g\right)\)

\(n_{CH_3COOH}=\dfrac{15}{60}=0,25\left(mol\right)\\ NaOH+CH_3COOH\rightarrow CH_3COONa+H_2O\\a, n_{CH_3COONa}=n_{CH_3COOH}=0,25\left(mol\right)\\ a=m_{CH_3COONa}=0,25.82=20,5\left(g\right)\\ b,m_{ddCH_3COOH}=\dfrac{15.100}{2}=750\left(g\right)\)

a, PT: \(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

b, Ta có: \(n_{Al}=\dfrac{5,4}{27}=0,2\left(mol\right)\)

Theo PT: \(\left\{{}\begin{matrix}n_{HCl}=3n_{Al}=0,6\left(mol\right)\\n_{H_2}=\dfrac{3}{2}n_{Al}=0,3\left(mol\right)\end{matrix}\right.\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(V_{H_2}=0,3.22,4=6,72\left(l\right)\)

c, Cách 1:

Theo PT: \(n_{AlCl_3}=n_{Al}=0,2\left(mol\right)\)

\(\Rightarrow m_{AlCl_3}=0,2.133,5=26,7\left(g\right)\)

Cách 2:

Ta có: \(m_{H_2}=0,3.2=0,6\left(g\right)\)

Theo ĐLBT KL, có: m_Al + m_HCl = m_AlCl3 + m_H2

⇒ m_AlCl3 = m_Al + m_HCl - m_H2 = 5,4 + 21,9 - 0,6 = 26,7 (g)

Bạn tham khảo nhé!

a, \(n_{Fe}=\dfrac{0,56}{56}=0,01\left(mol\right)\)

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

Theo PT: \(n_{FeSO_4}=n_{H_2}=n_{Fe}=0,01\left(mol\right)\)

\(\Rightarrow m_{FeSO_4}=0,01.152=1,52\left(g\right)\)

\(V_{H_2}=0,01.22,4=0,224\left(l\right)\)

b, \(n_{H_2SO_4}=n_{Fe}=0,01\left(mol\right)\Rightarrow m_{ddH_2SO_4}=\dfrac{0,01.98}{19,6\%}=5\left(g\right)\)

c, Ta có: m _{dd sau pư} = 0,56 + 5 - 0,01.2 = 5,54 (g)

\(\Rightarrow C\%_{FeSO_4}=\dfrac{1,52}{5,54}.100\%\approx27,44\%\)

Bài 1 :

PTHH : Zn + H₂SO₄ ------> ZnSO₄ + H₂

\(n_{Zn}=\frac{m}{M}=\frac{26}{65}=0,4\left(mol\right)\)

Theo PTHH : n_H2 = n_Zn = 0,4 mol

=> Khối lượng H₂ được tạo ra ở đktc là :

\(V=n\times22,4\)

\(\Rightarrow V_{H_2}=0,4\times22,4\)

\(\Rightarrow V_{H_2}=8,96\left(l\right)\)

Theo PTHH : n_ZnSO4 = n_Zn = 0,4 mol

=> Khối lượng muối được tạo thành là :

\(m=n\times M\)

\(\Rightarrow m_{ZnSO_4}=0,4\times161\)

\(\Rightarrow m_{ZnSO_4}=64,4\left(g\right)\)

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

=> Khối lượng \(H_2SO_4\)cần dùng cho phản ứng là :

\(m=n\times M\)

\(\Rightarrow m_{H_2SO_4}=0,4\times98\)

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

c) Nồng độ phần trăm thu được sau phản ứng là :

\(C\%=\frac{m_{ct}}{m_{dd}}\times100\%\)

\(\Rightarrow C\%=\frac{39,2}{64,4}\times100\%\approx60,9\%\)

Vậy :.........................

a)\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

b) Số mol của magie là:

\(n_{Mg}=\dfrac{m_{Mg}}{M_{Mg}}=\dfrac{3,6}{24}=0,15mol\)

\(PTHH:Mg+2HCl\xrightarrow[]{}MgCl_2+H_2\)

tỉ lệ       :1          2           1               1      mol

số mol  :0,15     0,3        0,15          0,15 mol

Thể tích khí hiđro là:

\(V_{H_2}=n_{H_2}.22,4=0,15.22,4=3,36\left(l\right)\)

c) Khối lượng muối tạo thành là:

\(m_{MgCl_2}=n_{MgCl_2}.M_{MgCl_2}=0,15.95=14,25\left(g\right)\)

d) Khối lượng axit đã dùng là:

\(m_{HCl}=n_{HCl}.M_{HCl}=0,3.36,5=10,95\left(g\right)\)