a, \(V_{C_2H_5OH}=\dfrac{10.9}{100}=0,9\left(l\right)=900\left(ml\right)\)

\(\Rightarrow m_{C_2H_5OH}=900.0,8=720\left(g\right)\Rightarrow n_{C_2H_5OH}=\dfrac{720}{46}=\dfrac{360}{23}\left(mol\right)\)

PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=\dfrac{360}{23}\left(mol\right)\)

Mà: H = 92%

\(\Rightarrow n_{CH_3COOH\left(TT\right)}=\dfrac{360}{23}.92\%=14,4\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH}=14,4.60=864\left(g\right)\)

b, \(m_{ddgiam}=\dfrac{864}{5\%}=17280\left(l\right)\)

Câu 8:

\(n_{C_2H_5OH}=\dfrac{4,6}{46}=0,1\left(mol\right)\)

PT: \(C_2H_5OH+O_2\underrightarrow{^{mengiam}}CH_3COOH+H_2O\)

Theo PT: \(n_{CH_3COOH\left(LT\right)}=n_{C_2H_5OH}=0,1\left(mol\right)\)

\(\Rightarrow m_{CH_3COOH\left(LT\right)}=0,1.60=6\left(g\right)\)

\(\Rightarrow H=\dfrac{4,8}{6}.100\%=80\%\)

Đáp án: A

B.40%

Khối lượng C có trong 1,5 tấn than là: \(m_C=\dfrac{1,5.95}{100}=1,425\) (g)

\(n_C=\dfrac{m}{M}=\dfrac{1,425}{12}=\dfrac{19}{160}\)(mol)

PTHH: C + O₂ → CO₂

Theo PT: \(n_{O_2}=n_C=\dfrac{19}{160}\)

Khối lượng khí oxi cần dùng là: \(m_{O_2}=n.M=\dfrac{19}{160}.32=3,8\)(g) 

Đổi: 200ml=0,2l

\(n_{Br_2}=C_M.V=0,5.0.2=0,1\) (mol)

PTHH: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)

\(\Rightarrow n_{C_2H_4}=n_{Br_2}=0,1\) (mol)

Thể tích C₂H₄ cần dùng là: \(V_{C_2H_4}=n.22,4=0,1.22,4=2,24\) (l)

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

b, \(n_P=\dfrac{6,2}{31}=0,2\left(mol\right)\)

\(n_{O_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

Xét tỉ lệ: \(\dfrac{0,2}{4}< \dfrac{0,4}{5}\), ta được O₂ dư.

Theo PT: \(n_{O_2\left(pư\right)}=\dfrac{5}{4}n_P=0,25\left(mol\right)\Rightarrow n_{O_2\left(dư\right)}=0,4-0,25=0,15\left(mol\right)\)

\(\Rightarrow m_{O_2\left(dư\right)}=0,15.32=4,8\left(g\right)\)

c, Theo PT: \(n_{P_2O_5}=\dfrac{1}{2}n_P=0,1\left(mol\right)\Rightarrow m_{P_2O_5}=0,1.142=14,2\left(g\right)\)

d, \(m_{P_2O_5}=14,2.80\%=11,36\left(g\right)\)

Giúp mik vs T-T

\(n_K=\dfrac{m}{M}=\dfrac{7,8}{39}=0,2\left(mol\right)\)

\(a,PTHH:4K+O_2\rightarrow2K_2O\)

                   \(0,2:0,05:0,1\left(mol\right)\)

\(K_2O+H_2O\rightarrow2KOH\)

\(0,1:0,1:0,2\left(mol\right)\)

\(b,V_{O_2}=n.22,4=0,05.22,4=1,12\left(l\right)\)

\(c,m_{KOH}=n.M=0,2.\left(39+16+1\right)=0,2.56=11,2\left(g\right)\)

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, \(n_{Mg}=\dfrac{1,2}{24}=0,05\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,1\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,1.36,5}{10\%}=36,5\left(g\right)\)

c, \(n_{H_2}=n_{MgCl_2}=n_{Mg}=0,05\left(mol\right)\)

Ta có: m _{dd sau pư} = 1,2 + 36,5 - 0,05.2 = 37,6 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,05.95}{37,6}.100\%\approx12,63\%\)

\(n_{Ag}=\dfrac{64,8}{108}=0,6\left(mol\right)\)

Gọi CTPT của X là C_nH_2nO

- TH1: X pư tạo 2 Ag.

\(\Rightarrow n_X=0,3\left(mol\right)\Rightarrow M_X=\dfrac{21,6}{0,3}=72\left(g/mol\right)\)

\(\Rightarrow12n+2n+16=72\Rightarrow n=4\)

Vậy: X là C₄H₈O.

- TH2: X pư tạo 4 Ag. ( X là HCHO, M = 30 g/mol)

\(\Rightarrow n_X=0,15\left(mol\right)\Rightarrow M_X=\dfrac{21,6}{0,15}=144\left(g/mol\right)\)

→ vô lý

Đáp án: D