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\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\\ a,PTHH:Mg+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Mg+H_2\\ n_{H_2}=n_{\left(CH_3COO\right)_2Mg}=n_{Mg}=0,2\left(mol\right);n_{CH_3COOH}=2.0,2=0,4\left(mol\right)\\ b,V_{H_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Mg}=142.0,2=28,4\left(g\right)\\ d,m_{ddCH_3COOH}=\dfrac{0,4.60.100}{12}=200\left(g\right)\)
\(n_{CaCO_3}=\dfrac{20}{100}=0,2\left(mol\right)\\ a,PTHH:CaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ca+CO_2+H_2O\\ b,n_{CO_2}=n_{\left(CH_3COO\right)_2Ca}=n_{CaCO_3}=0,2\left(mol\right)\\ b,V_{CO_2\left(đktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Ca}=0,2.158=31,6\left(g\right)\\ d,C_4H_{10}+\dfrac{5}{2}O_2\rightarrow2CH_3COOH+H_2O\\ n_{C_4H_{10}\left(LT\right)}=\dfrac{0,4}{2}=0,2\left(mol\right)\\ n_{C_4H_{10}\left(TT\right)}=0,2:50\%=0,4\left(mol\right)\\ m_{C_4H_{10}\left(tt\right)}=58.0,4=23,2\left(g\right)\)
\(n_{BaCO_3}=\dfrac{19,7}{197}=0,1\left(mol\right)\\ a,PTHH:BaCO_3+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Ba+CO_2+H_2O\\ n_{CO_2}=n_{\left(CH_3COO\right)_2Ba}=n_{BaCO_3}=0,1\left(mol\right)\\ b,V_{CO_2\left(đktc\right)}=0,1.22,4=2,24\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Ba}=255.0,1=25,5\left(g\right)\\ d,C_2H_5OH+O_2\rightarrow\left(men.giấm\right)CH_3COOH+H_2O\\ n_{CH_3COOH}=0,1.2=0,2\left(mol\right);n_{C_2H_5OH\left(LT\right)}=n_{CH_3COOH}=0,2\left(mol\right)\\ n_{C_2H_5OH\left(TT\right)}=0,2.75\%=0,15\left(mol\right)\\ m_{C_2H_5OH\left(TT\right)}=0,15.46=6,9\left(g\right)\)
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: \(m_{dd}=m_{Mg}+m_{ddHCl}-m_{H_2}=53,3\left(g\right)\)
\(\Rightarrow m_{ddHCl}=50\left(g\right)\) \(\Rightarrow C\%_{HCl\left(p.ứ\right)}=\dfrac{10,95}{50}\cdot100\%=21,9\%\)
\(n_{HCl}=\dfrac{18.25}{36.5}=0.5\left(mol\right)\)
\(a.Mg+2HCl\rightarrow MgCl_2+H_2\)
\(b.\)
\(n_{Mg}=n_{H_2}=\dfrac{1}{2}\cdot n_{HCl}=\dfrac{1}{2}\cdot0.5=0.25\left(mol\right)\)
\(m_{Mg}=0.25\cdot24=6\left(g\right)\)
\(V_{H_2}=0.25\cdot22.4=5.6\left(l\right)\)
\(c.\)
\(V_{H_2\left(tt\right)}=5.6\cdot90\%=5.04\left(l\right)\)
\(Zn + H_2SO_4 \rightarrow ZnSO_4 + H_2\)
b)
\(n_{H_2}= \dfrac{2,24}{22,4}= 0,1 mol\)
\(\)Theo PTHH:
\(n_{ZnSO_4}= n_{H_2}= 0,1 mol\)
\(m_{ZnSO_4}= 0,1 . 161=16,1g\)
c)
Theo PTHH:
\(n_{H_2SO_4}= n_{H_2}= 0,1 mol\)
\(\Rightarrow m_{H_2SO_4}= 0,1 . 98= 9,8g\)
\(\Rightarrow m_{dd H_2SO_4}= \dfrac{9,8 . 100}{20}=49g\)
Ta có: \(n_{Fe}=\dfrac{28}{56}=0,5\left(mol\right)\)
a, PT: \(Fe+H_2SO_4\rightarrow FeSO_4+H_2\)
______0,5____0,5_____0,5_____0,5 (mol)
b, mH2SO4 = 0,5.98 = 49 (g)
c, mFeSO4 = 0,5.152 = 76 (g)
d, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
____0,5__0,5 (mol)
⇒ mCu = 0,5.64 = 32 (g)
Bạn tham khảo nhé!
PTHH: \(NaOH+HCl\rightarrow NaCl+H_2O\)
a_____________a (mol)
\(KOH+HCl\rightarrow KCl+H_2O\)
b_____________b
Ta lập HPT: \(\left\{{}\begin{matrix}40a+56b=3,04\\58,5a+74,5b=4,15\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=0,02\\b=0,04\end{matrix}\right.\)
\(\Rightarrow...\)
\(n_{Zn}=\dfrac{19,5}{65}=0,3\left(mol\right)\\ a,PTHH:Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\\ b,n_{H_2}=n_{\left(CH_3COO\right)_2Zn}=n_{Zn}=0,3\left(mol\right);n_{CH_3COOH}=2.0,3=0,6\left(mol\right)\\ b,V_{H_2\left(đkc\right)}=24,79.0,3=7,437\left(l\right)\\ c,m_{\left(CH_3COO\right)_2Zn}=0,3.183=54,9\left(g\right)\\ d,C_2H_5OH+CH_3COOH\rightarrow CH_3COOC_2H_5+H_2O\\ n_{Este\left(LT\right)}=n_{CH_3COOH}=0,6\left(mol\right)\\ n_{este\left(TT\right)}=80\%.0,6=0,48\left(mol\right)\\ m=m_{este\left(TT\right)}=88.0,48=42,24\left(g\right)\)