a) Khối lượng rượu etylic:

n_CO2 = \(\dfrac{11.2}{22.4}\) = 0.5 mol

Phương trình lên men glucozơ:

C6H12O6 → 2C2H5OH + 2CO2.

Theo pt: nrượu etylic = nCO2 = 0,5 mol.

mrượu etylic = 0,5 × 46 = 23g.

b) Khối lượng glucozơ.

Theo phương trình n_glucozo = 1/2. nCO2 = 1/2 . 0,5 = 0,25 mol

Do hiệu suất đạt 90% nên khối lượng glucozo cần dùng là:  0,25.180.\(\dfrac{100}{90}\)= 50 (g)

a, \(n_{CO_2}=\dfrac{1,7353}{24,79}=0,07\left(mol\right);n_{NaOH}=\dfrac{6,4}{40}=0,16\left(mol\right)\)

PTHH: CO₂ + 2NaOH → Na₂CO₃ + H₂O

Mol:      0,07      0,07          0,07

Ta có: \(\dfrac{0,07}{1}< \dfrac{0,16}{2}\) ⇒ CO₂ hết, NaOH dư

b, \(m_{Na_2CO_3}=0,07.106=7,42\left(g\right)\)

c, \(m_{NaOHdư}=\left(0,16-0,07\right).40=3,6\left(g\right)\)

 \(\Rightarrow m_{muối}=7,42+3,6=11,02\left(g\right)\)

tks bạn nha

TK

Từ C₂H₄O₂ ta có: M = 60 g/mol; m_C = 2 x 12 = 24 g; m_H = 4 x 1 = 4 g;

M_O = 2 x 16 = 32 g.

%C = (24 : 60) x 100% = 40%; %H = (4 : 60) x 100% = 6,67%;

%O = 100% - 40% - 6,67% = 53,33%.

a, \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\)

b, \(n_{Mg}=\dfrac{4,1}{24}=\dfrac{41}{240}\left(mol\right)\)

Theo PT: \(n_{H_2}=n_{MgSO_4}=n_{Mg}=\dfrac{41}{240}\left(mol\right)\)

\(\Rightarrow V_{H_2}=\dfrac{41}{240}.24,79\approx4,23\left(l\right)\)

c, \(m_{MgSO_4}=\dfrac{41}{240}.120=20,5\left(g\right)\)

\(n_{Mg}=\dfrac{4,1}{24}=\dfrac{41}{210}\left(mol\right)\)

PTHH :

\(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)

41/210                 41/210      41/210

\(b,V_{H_2}=\dfrac{41}{240}.24,79=4,235\left(l\right)\)

\(c,m_{MgSO_4}=\dfrac{41}{240}.120=20,5\left(g\right)\)

\(n_{Fe}=\dfrac{m}{M}=\dfrac{11,2}{56}=0,2\left(mol\right)\\ PTHH:Fe+H_2SO_4\rightarrow FeSO_4+H_2\uparrow\\ \left(mol\right)-0,2-\rightarrow0,2---0,2--0,2\\ m_{FeSO_4}=n.M=0,2.152=30,4\left(g\right)\\ V_{H_2}=n.22,4=0,2.22,4=2,24\left(l\right)\)

a, \(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)

Ta có: \(n_{C_2H_4}=\dfrac{11,2}{22,4}=0,5\left(mol\right)\)

Theo PT: \(n_{O_2}=3n_{C_2H_4}=1,5\left(mol\right)\Rightarrow V_{O_2}=1,5.22,4=33,6\left(l\right)\)

b, \(V_{kk}=\dfrac{V_{O_2}}{20\%}=168\left(l\right)\)