Bài 9:

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\\ n_{HCl}=\dfrac{14,6\%.100}{36,5}=0,4\left(mol\right)\\ PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ Vì:\dfrac{0,1}{1}< \dfrac{0,4}{2}\Rightarrow HCldư\\ n_{HCl\left(dư\right)}=0,4-2.0,1=0,2\left(mol\right)\\ m_{HCl\left(dư\right)}=0,2.36,5=7,3\left(g\right)\)

Bài 10:

\(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\\ PTHH:CaO+H_2O\rightarrow Ca\left(OH\right)_2\\ n_{Ca\left(OH\right)_2}=n_{CaO}=0,1\left(mol\right)\\ m_{Ca\left(OH\right)_2}=0,1.74=7,4\left(g\right)\\ m_{ddCa\left(OH\right)_2}=m_{CaO}+m_{H_2O}=5,6+500.1=505,6\left(g\right)\\ a,C\%_{ddCa\left(OH\right)_2}=\dfrac{7,4}{505,6}.100\approx1,464\%\\ b,V_{ddCa\left(OH\right)_2}=V_{H_2O}=500\left(ml\right)=0,5\left(l\right)\\ C_{MddCa\left(OH\right)_2}=\dfrac{0,1}{0,5}=0,2\left(M\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1mol\)

\(m_{HCl}=\dfrac{200\cdot14,6\%}{100\%}=29,2g\Rightarrow n_{HCl}=0,8mol\)

a)\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

   0,1      0,8            0          0

   0,1      0,2            0,1       0,1

   0         0,6            0,1       0,1

b)Chất HCl dư và dư \(m=0,6\cdot36,5=21,9g\)

c)\(V_{H_2}=0,1\cdot22,4=2,24l\) 

d)\(m_{H_2}=0,1\cdot2=0,2g\)

\(m_{ZnCl_2}=0,1\cdot136=13,6g\)

\(m_{ddZnCl_2}=6,5+200-0,2=206,3g\)

\(C\%=\dfrac{13,6}{206,3}\cdot100\%=6,59\%\)

a, ta có pt sau : Zn + 2HCl >ZnCl₂ + H₂ (1)

b,  n_HCl=\(\dfrac{200\times14,6}{100}=29,2\left(g\right)\)\(\Rightarrow n_{HCl}=\dfrac{29,2}{36,5}=0,8\left(mol\right)\)

Ta có : n_Zn=\(\dfrac{6,5}{65}=0,1\left(mol\right)\)

Ta có tỉ lệ số mol là : _{\(\dfrac{n_{Zn}}{1}< \dfrac{n_{HCl}}{2}\left(\dfrac{0,1}{1}< \dfrac{0,8}{2}\right)\)}

\(\Rightarrow\) HCl dư , Zn pứ hết 

Theo pt : n_HClpứ = 2.n_Zn=2.0,1=0,2(mol) 

\(\Rightarrow\)n_{HCl dư} = n_{HCl bđ} - n_{HCl pứ} = 0,8 - 0,2 = 0,6 (mol) 

\(\Rightarrow\)m_{HCl dư}=0,6.36,6=21,9 (g)

c,theo pt :n_H2=n_Zn=0,1(mol)

_{\(\Rightarrow\)}V_H2=0,1.22,4=2,24(l)

d,Các chất có trong dung dịch sau pứ là: ZnCl₂ , HCl dư 

mk chịu câu này 

                  \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

trc p/u :      0,1      0,4         

p/u :           0,1      0,2         0,1           0,1

Sau :          0        0,2           0,1        0,1 

a, ----> Sau p/u HCl dư 

\(m_{HCldư}=0,2.26,5=7,3\left(g\right)\)

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(m_{HCl}=\dfrac{200.7,3}{100}=14,6\left(g\right)\)

\(n_{HCl}=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

b, \(m_{ZnCl_2}=0,1.136=13,6\left(g\right)\)

\(m_{ddZnCl_2}=6,5+200-\left(0,1.2\right)=206,3\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{13,6}{206,3}.100\%\approx6,59\%\)

c, \(V_{H_2}=0,1.22,4=2,24\left(l\right)\)

Thể tích H2 thu được thực tế :

\(V_{H_2tt}=2,24.90\%=2,016\left(l\right)\)

PTHH: \(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

a) Ta có: \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\) \(\Rightarrow n_{HCl}=0,2mol\)

\(\Rightarrow m_{ddHCl}=\dfrac{0,2\cdot36,5}{10,95\%}\approx66,67\left(g\right)\)

b) Theo PTHH: \(n_{ZnCl_2}=n_{H_2}=n_{Zn}=0,1mol\)

\(\Rightarrow\left\{{}\begin{matrix}m_{ZnCl_2}=0,1\cdot136=13,6\left(g\right)\\m_{H_2}=0,1\cdot2=0,2\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{Zn}+m_{ddHCl}-m_{H_2}=72,97\left(g\right)\)

\(\Rightarrow C\%_{ZnCl_2}=\dfrac{13,6}{72,97}\cdot100\%\approx18,64\%\)

Zn+2HCl->ZnCl2+H2

0,1---0,2

n Zn=0,1 mol

=>HCl dư

\(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\uparrow\)

0,1                     0,1         0,1 

\(b,C_M=\dfrac{0,1}{0,1}=1M\)

\(c,V_{H_2}=0,1.22,4=2,24\left(l\right)\)

PTHH: \(K_2SO_3+2HCl\rightarrow2KCl+H_2O+SO_2\uparrow\)

a+b+c) Ta có: \(n_{SO_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{KCl}=0,5\left(mol\right)=n_{HCl}\\n_{K_2SO_3}=0,25\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_3}=0,25\cdot158=39,5\left(g\right)=a\\m_{KCl}=0,5\cdot74,5=37,25\left(g\right)\\m_{ddHCl}=\dfrac{0,5\cdot36,5}{10,95\%}\approx166,67\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{SO_2}=0,25\cdot64=16\left(g\right)\)

\(\Rightarrow m_{dd}=m_{K_2SO_3}+m_{ddHCl}-m_{SO_2}=190,17\left(g\right)\) \(\Rightarrow C\%_{KCl}=\dfrac{37,25}{190,17}\cdot100\%\approx19,59\%\)

d) PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)

Theo PTHH: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)

Ta có: \(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\)

\(n_{HCl}=4.100:1000=0,4\left(mol\right)\)

a. PTHH: MgO + 2HCl ---> MgCl₂ + H₂O

Ta thấy: \(\dfrac{0,1}{1}< \dfrac{0,4}{2}\)

Vậy HCl dư.

=> \(n_{dư}=\dfrac{0,1.2}{0,4}=0,5\left(mol\right)\)

=> \(m_{dư}=0,5.36,5=18,2\left(g\right)\) 

b. Ta có: \(V_{dd_{MgCl_2}}=V_{HCl}=\dfrac{100}{1000}=0,1\left(lít\right)\)

Theo PT: \(n_{MgCl_2}=n_{MgO}=0,1\left(mol\right)\)

=> \(C_{M_{MgCl_2}}=\dfrac{0,1}{0,1}=1M\)

Số mol Zn: \(n_{Zn}=\frac{6,5}{65}=0,1\left(mol\right)\)

Khối lượng \(CH_3COOH\): \(m_{CH_3COOH}=\frac{100\cdot10}{100}=10\left(g\right)\)

Số mol \(CH_3COOH\): \(n_{CH_3COOH}=\frac{10}{60}=0,6\left(mol\right)\)

PTHH:

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\uparrow\)(1)

0,1 0,6 (mol)

Vì \(\frac{0,1}{1}< \frac{0,6}{2}\) nên \(CH_3COOH\) dư lượng chất được tính theo \(Zn\).

Từ PTHH(1): \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,1\left(mol\right)\)

Khối lượng của \(\left(CH_3COO\right)_2Zn\): \(m_{\left(CH_3COO\right)_2Zn}=0,1\cdot183=18,3\left(g\right)\)

Nồng độ phần trong chất thu được sau phản ứng:

\(C\%_{\left(CH_3COO\right)_2Zn}=\frac{18,3}{100}\cdot100\%=18,3\%\)

Mink ko chắc lắm nha bạn vì mink ko học giỏi về môn này

nZn= 0.1 mol

mCH3COOH= 10g

nCH3COOH= 1/6 mol

2CH3COOH + Zn --> (CH3COO)2Zn + H2

Từ PTHH:

1/6 :2 > 0.1 =>Zn dư

n(CH3COO)Zn= 1/12 mol

nH2= 1/12 mol

mH2= 1/6 g

mdd sau phản ứng=6.5+100-1/6=319/3g

C%(CH3COO)2Zn= 15.25/319/3 *100%= 14.34%

\(n_{HCl}=\dfrac{100.14,6\%}{36,5}=0,4\left(mol\right)\\ n_{MgCO_3}=\dfrac{50}{84}=\dfrac{25}{42}\left(mol\right)\\ PTHH:MgCO_3+2HCl\rightarrow MgCl_2+CO_2+H_2O\\ Vì:0,4:2< \dfrac{25}{42}:1\\ \Rightarrow MgCO_3dư\\ \Rightarrow ddsau:MgCl_2\\n_{MgCO_3\left(p.ứ\right)}=n_{CO_2}= n_{MgCl_2}=\dfrac{n_{HCl}}{2}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ m_{ddsau}=m_{MgCO_3\left(p.ứ\right)}+m_{ddHCl}-m_{CO_2}=0,2.84+100-0,2.44=108\left(g\right)\\ C\%_{ddMgCl_2}=\dfrac{0,2.95}{108}.100\approx17,593\%\%\)