Đề kiểm tra cuối kì II: đề 1

Gọi số mol NaCl và NaBr lần lượt là x, y.

\(\Rightarrow58,5x+103y=26,45\)

\(NaCl+AgNO_3\rightarrow AgCl+NaNO_3\)

\(NaBr+AgNO_3\rightarrow NaNO_3+AgBr\)

\(n_{AgCl}=n_{NaCl}=x\)

\(n_{AgBr}=n_{NaBr}=y\)

\(\Rightarrow m_{kt}=m_{AgCl}+m_{AgBr}=143,5x+188y=51,95\)

\(\Rightarrow\left\{{}\begin{matrix}x=0,1\\y=0,2\end{matrix}\right.\)

\(\Rightarrow m_{NaCl}=58,5x=5,85\left(g\right)\)

\(m_{NaBr}=103y=20,6\left(g\right)\)

\(n_{AgNO3}=n_{AgCl}+n_{AgBr}=x+y=0,3\left(mol\right)=a\)

\(n_{CO2}=\frac{13,2}{44}=0,3\left(mol\right)\)

\(n_{H2O}=\frac{5,4}{18}=0,3\left(mol\right)\)

\(n_{H2O}=n_{CO2}\Rightarrow A:anken\)

\(C_nH_{2n}+\frac{3n}{2}O_2\rightarrow nCO_2+H_2O\)

\(\Rightarrow n_C:n_H=0,3:0,6=1:2\)

CTĐGN : CH2

\(M_A=28\Rightarrow14n=28\Rightarrow n=2\)

Vậy CTPT là C2H4

Ta có :

\(n_{CO2}=\frac{4,48}{22,4}=0,2\left(mol\right)\)

\(n_{NaOH}=0,5.1=0,5\left(mol\right)\)

\(\frac{n_{NaOH}}{n_{CO2}}=\frac{0,5}{0,2}=2,5\)

\(\Rightarrow Na_2CO_3\)

\(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

\(n_{Na2CO3}=n_{CO2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Na2CO3}=0,2.106=21,2\left(g\right)\)

\(n_{NaOH}=0,5-0,4=0,1\left(mol\right)\)

\(\Rightarrow CM_{Na2CO3}=\frac{0,2}{0,5}=0,4M\)

\(\Rightarrow CM_{NaOH}=\frac{0,1}{0,5}=0,2M\)

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

\(n_{NaOH}=0,5.1=0,5\left(mol\right)\)

Ta có: \(T=\frac{n_{NaOH}}{n_{CO_2}}=\frac{0,5}{0,2}=2,5>2\)

=> Chỉ tạo muối Na2CO3

PTHH: \(2NaOH+CO_2\rightarrow Na_2CO_3+H_2O\)

Theo PT:...2.............1.......

Theo đề:....5.............0,2....

Lập tỉ lệ : \(\frac{0,5}{2}>\frac{0,2}{1}\Rightarrow\)NaOH dư

Theo PT: \(n_{Na_2CO_3}=n_{CO_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Na_2CO_3}=0,2.106=21,2\left(g\right)\)

b)Dd au phản ứng gồm: NaOH dư, Na2CO3

\(n_{NaOH\left(dư\right)}=0,5-0,2.2=0,1\left(mol\right)\)

\(\Rightarrow CM_{NaOH}=\frac{0,1}{0,5}=0,2M\)

\(\Rightarrow CM_{Na_2CO_3}=\frac{0,2}{0,5}=0,4M\)

Ta có:

\(n_{CO2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

\(n_{NaOH}=0,4.1=0,4\left(mol\right)\)

\(\frac{n_{NaOH}}{n_{CO2}}=\frac{0,4}{0,5}=0,8\)

\(NaOH+CO_2\rightarrow NaHCO_3\)

\(n_{NaHCO3}=n_{NaOH}=0,4\left(mol\right)\)

\(\Rightarrow m_{NaHCO3}=0,4.84=33,6\left(g\right)\)

\(\Rightarrow CM_{NaHCO3}=\frac{0,4}{0,4}=1M\)

\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

Ta có:

\(n_{CH4}=\frac{6,72}{22,4}=0,3\left(mol\right)\)

\(\Rightarrow n_{CO2}=n_{CH4}=0,3\left(mol\right)\)

\(\Rightarrow n_{CaCO3}=n_{CO2}=0,3\left(mol\right)\)

\(n_{H2O}=0,6\left(mol\right)\)

\(\Rightarrow m_1=m_{CO2}+m_{H2O}=0,3.44+0,6.18=24\left(g\right)\)

\(\Rightarrow m_2=m_{CaCO3}=0,3.100=30\left(g\right)\)

\(CH_4+2O_2\rightarrow CO_2+2H_2O\)

\(C_2H_4+3O_2\rightarrow2CO_2+2H_2O\)

Ta có:

\(n_{CO2}=\frac{11,2}{22,4}=0,5\left(mol\right)\)

Gọi số mol của CH4 là a , số mol của C2H4 là b

Giải hệ PT:

\(\left\{{}\begin{matrix}16a+28b=7,2\\a+2n=0,5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=0,1\\b=0,2\end{matrix}\right.\)

\(\%m_{CH4}=\frac{0,1.16}{7,2}.100\%=22,22\%\)

\(\%m_{C2H4}=100\%-22,22\%=77,78\%\)

a. \(Cl_2+H_2\rightarrow2HCl\)

b.\(2Al+7O_2+6H_2O\rightarrow2AlIO_3+12HI\)

c.\(Cl_2+Ca\left(OH\right)_2\rightarrow CaOCl_2+H_2O\)

d. \(Br_2+H_2O\rightarrow HBr+HBrO\)

e. \(SiO_2+4HF\rightarrow SiF_4+2H_2O\)

f.\(HCl+NaHCO_3\rightarrow NaCl+H_2O+CO_2\)

\(\overline{M}=\frac{40}{3}.2=\frac{80}{3}\)

\(\left\{{}\begin{matrix}n_{CH4}:a\left(mol\right)\\n_{O2}:b\left(mol\right)\end{matrix}\right.\Rightarrow\frac{16a+32b}{a+b}=\frac{80}{3}\)

\(\Leftrightarrow\frac{32}{3}a=\frac{16}{3}b\Leftrightarrow\frac{a}{b}=\frac{1}{2}\)

Giả sử \(\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right.\)

\(CH_4+2O_2\underrightarrow{^{to}}CO_2+H_2O\)

1_____2____________________

\(\Rightarrow\) Phản ứng hoàn toàn

\(MC_nH_{2n+1}Br=75,5.2=151\)

\(\Rightarrow M_{CnH2n+2}=72\)

\(\Rightarrow14n+2=72\)

\(\Rightarrow n=5\)

Vậy công thức phân tử là C5H12