\(n_{MgCO_3}=\dfrac{12.6}{84}=0.15\left(mol\right)\)

\(Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

\(0.15..........0.3............................................0.15\)

\(V_{CO_2}=0.15\cdot22.4=3.36\left(l\right)\)

\(m_{CH_3COOH}=0.3\cdot60=18\left(g\right)\)

\(C\%_{CH_3COOH}=\dfrac{18}{200}\cdot100\%=9\%\)

Pthh MgCO3+2CH3COOH--->(CH3COO)2Mg+CO2+H2O

Ta có nMgCO3=0,15 mol

Theo pthh thì nCH3COOH=0,3 mol

=>mCH3COOH=17,7 g

=>C%CH3COOH=8,85%

Câu a mik chưa bt tính V của cái nào

a, \(n_{CO_2}=\dfrac{2,479}{24,79}=0,1\left(mol\right)\)

PT: \(MgO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2O\)

\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

Theo PT: \(n_{MgCO_3}=n_{CO_2}=0,1\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{0,1.84}{10,4}.100\%\approx80,77\%\\\%m_{MgO}\approx19,23\%\end{matrix}\right.\)

b, \(n_{MgO}=\dfrac{10,4-0,1.84}{40}=0,05\left(mol\right)\)

Theo PT: \(n_{CH_3COOH}=2n_{MgO}+2n_{MgCO_3}=0,3\left(mol\right)\)

\(\Rightarrow C_{M_{CH_3COOH}}=\dfrac{0,3}{0,2}=1,5\left(M\right)\)

Hơi muộn nhé, lần sau bạn đăng sớm hơn chứ hầu hết giờ nay mọi người đi ngủ rồi, chúc bạn thi tốt nhé ^^

a) Gọi \(\left\{{}\begin{matrix}n_{MgCO_3}=a\left(mol\right)\\n_{CaCO_3}=b\left(mol\right)\end{matrix}\right.\)

\(n_{CO_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

PTHH:

MgCO₃ + 2CH₃COOH ---> (CH₃COO)₂Mg + CO₂ + H₂O

a---------->2a-------------------->a----------------->a

CaCO₃ + 2CH₃COOH ---> (CH₃COO)₂Ca + CO₂ + H₂O

b---------->2b------------------>b------------------>b

=> \(\left\{{}\begin{matrix}84a+100b=31,8\\a+b=0,35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=0,2\left(mol\right)\\b=0,15\left(mol\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m_{MgCO_3}=0,2.84=16,8\left(g\right)\\m_{CaCO_3}=0,15.100=15\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\%m_{MgCO_3}=\dfrac{16,8}{31,8}.100\%=52,8\%\\\%m_{CaCO_3}=100\%-52,8\%=47,2\%\end{matrix}\right.\)

b) \(V_{dd}=\dfrac{2.0,2+2.0,15}{2}=0,35\left(l\right)\)

c) \(m_{muối}=0,2.142+0,15.158=52,1\left(g\right)\)

Câu 8 : 

\(n_{MgCO3}=\dfrac{42}{84}=0,5\left(mol\right)\)

Pt : \(2CH_3COOH+MgCO_3\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

                 1                  0,5                                         0,5

a) \(V_{CO2\left(dktc\right)}=0,5.22,4=11,2\left(l\right)\)

b) \(V_{CH3COOH}=\dfrac{1}{2}=0,5\left(l\right)\)

c) Pt : \(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

                  1                                         1

300ml = 0,3l

\(C_{MCH3COONa}=\dfrac{1}{0,3}=\dfrac{10}{3}\left(M\right)\)

 Chúc bạn học tốt

Xem lại đề nhé !

Gọi \(\left\{{}\begin{matrix}n_{Mg}:x\left(mol\right)\\n_{MgCO3}:y\left(mol\right)\end{matrix}\right.\)

\(2CH_3COOH+Mg\rightarrow\left(CH_3COO\right)_2Mg+H_2\)

\(MgCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+CO_2+H_2O\)

\(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)

Ta có:

\(n_{CaCO3}=n_{CO2}=y=\frac{5}{100}=0,05\left(mol\right)\)

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

\(\Rightarrow m=0,2.24+0,05.120=10,8\left(g\right)\)

Câu 1: D
Câu 3: B
Câu 8: B

Câu 9: A

Câu 10: B

Câu 15: D

Câu 21: C

Câu 23: B
Câu 2: 

a) \(m_{CH_4}=n.M=0,03.16=0,48\left(g\right)\)

b) \(m_{C\text{uS}O_4}=n.M=0,5.160=80\left(g\right)\)

c) \(n_{CO}=\dfrac{V_{\left(\text{đ}ktc\right)}}{22,4}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)

`=>` \(m_{CO}=n.M=0,1.28=2,8\left(g\right)\)

Câu 3:
a) \(V_{C_2H_4\left(\text{đ}ktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)

b) \(V_{CO_2\left(\text{đ}ktc\right)}=n.M=3.22,4=67,2\left(l\right)\)

c) \(n_{O_2}=\dfrac{m}{M}=\dfrac{1,6}{32}=0,05\left(mol\right)\)

`=>` \(V_{O_2\left(\text{đ}ktc\right)}=n.22,4=0,05.22,4=1,12\left(l\right)\)

d) \(n_{H_2S}=\dfrac{m}{M}=\dfrac{6,8}{34}=0,2\left(mol\right)\)

`=>` \(V_{H_2S\left(\text{đ}ktc\right)}=n.22,4=0,2.22,4=4,48\left(l\right)\)

AI GIÚP LUÔN VỚI ĐC HOK Ạ
ĐANG CẦN GẤP LẮM Ạ

nMgCO3 = 8.4/84 = 0.1 mol 

nCO2 = 0.05 mol 

MgCO3 + 2HCl => MgCl2 + CO2 + H2O 

0.05______0.1___________0.05 

nMgCO3= 0.05 < 0.1 => MgCO3 dư 

mMgCO3 dư = 0.05*84 = 4.2 g 

mHCl = 0.1*36.5 = 3.65 g 

Đáp án B.

3.

CH₃COOH + Na₂CO₃  →

CH₃COOH + CaCO₃  →

CH₃COOH + KHCO₃  →