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\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)
Câu 4:
a, \(n_{H_2}=\dfrac{5,6}{22,4}=0,25\left(mol\right)\)
PT: \(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
Theo PT: \(n_{CH_3COOH}=2n_{H_2}=0,5\left(mol\right)\Rightarrow V_{CH_3COOH}=\dfrac{0,5}{2,5}=0,2\left(l\right)\)
b, \(n_{\left(CH_3COO\right)_2Zn}=n_{H_2}=0,25\left(mol\right)\)
\(\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,25.183=45,75\left(g\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\\ b.n_{H_2}=0,2\left(mol\right)\\ TheoPT:n_{Mg}=n_{H_2}=0,2\left(mol\right)\\ \Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\\ c.n_{HCl}=2n_{H_2}=0,4\left(mol\right)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6\left(g\right)\)
a, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)
PT: \(2CH_3COOH+Zn\rightarrow\left(CH_3COO\right)_2Zn+H_2\)
Theo PT: \(n_{H_2}=n_{Zn}=0,1\left(mol\right)\Rightarrow V_{H_2}=0,1.22,4=2,24\left(l\right)\)
b, \(n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,1\left(mol\right)\Rightarrow m_{\left(CH_3COO\right)_2Zn}=0,1.183=18,3\left(g\right)\)
c, \(n_{CH_3COOH}=2n_{Zn}=0,2\left(mol\right)\)
\(\Rightarrow m_{CH_3COOH}=0,2.60=12\left(g\right)\)
\(\Rightarrow m_{ddCH_3COOH}=\dfrac{12}{60\%}=20\left(g\right)\)
a) PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)
b) Ta có: \(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)=n_{MgCl_2}\)
\(\Rightarrow m_{MgCl_2}=0,15\cdot95=14,25\left(g\right)\)
c) Theo PTHH: \(\left\{{}\begin{matrix}n_{HCl\left(p.ứ\right)}=0,3\left(mol\right)\\n_{H_2}=0,15\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{HCl\left(p.ứ\right)}=0,3\cdot36,5=10,95\left(g\right)\\m_{H_2}=0,15\cdot2=0,3\left(g\right)\end{matrix}\right.\)
Mặt khác: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
Ta có: \(n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\)
a. PTHH:
\(Zn+H_2SO_4--->ZnSO_4+H_2\)
\(Cu+H_2SO_4--\times-->\)
b. Theo PT: \(n_{Zn}=n_{H_2}=0,1\left(mol\right)\)
\(\Rightarrow m_{Zn}=0,1.65=6,5\left(g\right)\)
\(\Rightarrow m_{chất.rắn.còn.lại.sau.PỨ}=m_{Cu}=10,5-6,5=4\left(g\right)\)
c. Theo PT: \(n_{H_2SO_4}=n_{Zn}=0,1\left(mol\right)\)
\(\Rightarrow m_{H_2SO_4}=0,1.98=9,8\left(g\right)\)
Ta có: \(C_{\%_{H_2SO_4}}=\dfrac{9,8}{m_{dd_{H_2SO_4}}}.100\%=20\%\)
\(\Leftrightarrow m_{dd_{H_2SO_4}}=49\left(g\right)\)
\(a,PTHH:Zn+H_2SO_4\rightarrow ZnSO_4+H_2\\ b,m_{\text{chất rắn sau p/ứ}}=m_{Cu}\\ n_{H_2}=\dfrac{2,24}{22,4}=0,1\left(mol\right)\\ \Rightarrow n_{Zn}=0,1\left(mol\right)\\ \Rightarrow m_{Zn}=0,1\cdot65=6,5\left(g\right)\\ \Rightarrow m_{Cu}=10,5-6,5=4\left(g\right)\\ c,n_{H_2SO_4}=0,1\left(mol\right)\\ \Rightarrow m_{CT_{H_2SO_4}}=0,1\cdot98=9,8\left(g\right)\\ \Rightarrow m_{dd_{H_2SO_4}}=\dfrac{9,8\cdot100\%}{20\%}=49\left(g\right)\)
\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2|\)
1 1 1 1
0,05 0,05
b) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
⇒ \(m_{Mg}=0,05.24=1,2\left(g\right)\)
Chúc bạn học tốt
\(a,PTHH:Zn+2HCl\rightarrow ZnCl_2+H_2\\ b,n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ n_{HCl}=2\cdot0,15=0,3\left(mol\right)\)
Vì \(\dfrac{n_{Zn}}{1}>\dfrac{n_{HCl}}{2}\) nên sau p/ứ Zn dư
\(\Rightarrow n_{Zn}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\\ \Rightarrow m_{Zn}=0,15\cdot65=9,75\\ \Rightarrow m_{Zn\left(dư\right)}=13-9,75=3,25\left(g\right)\\ c,n_{H_2}=n_{Zn}=0,15\left(mol\right)\\ \Rightarrow V_{H_2}=0,15\cdot22,4=3,36\left(l\right)\)
a/ \(n_{H_2}=\dfrac{7,437}{24,79}=0,3\left(mol\right)\)
PTHH: Mg + 2HCl → MgCl2 + H2
Mol: 0,3 0,6 0,3 0,3
\(m_{Mg}=0,3.24=7,2\left(g\right)\)
b/ \(m_{MgCl_2}=0,3.95=28,5\left(g\right)\)
c/ \(m_{HCl}=0,3.36,5=10,95\left(g\right)\)
a)
2CH3COOH + Mg → (CH3COO)2Mg + H2
b)
nH2 = 0,448 : 22,4 = 0,02 mol
=> n(CH3COO)2Mg = 0,02 mol
<=> m(CH3COO)2Mg = 0,02.142 = 2,84 gam.
c) nCH3COOH = 2 nH2 = 0,04 mol
<=> mCH3COOH = 0,04 . 60 = 2,4 gam
=> C%CH3COOH = \(\dfrac{2,4}{40}.100\%\)= 6%