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MgCO3+2HCl->MgCl2+H2O+CO2
0,25--------0,5
n MgCO3=\(\dfrac{21}{84}\)=0,25 mol
=>VHCl=\(\dfrac{0,5}{2}\)=0,25 l=250ml
=>B
\(n_{MgCO_3}=\dfrac{12,6}{84}=0,15\left(mol\right)\)
\(MgCO_3+2HCl\rightarrow MgCl_2+H_2O+CO_2\)
0,15 0,15
\(V_{CO_2}=0,15.22,4=3,36\left(l\right)\)
--> A
\(n_{Zn}=\dfrac{26}{65}=0,4\left(mol\right)\)
a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)
1 2 1 1
0,4 0,8 0,4
b) \(n_{H2}=\dfrac{0,4.1}{1}=0,4\left(mol\right)\)
\(V_{H2\left(dtkc\right)}=0,4.22,4=8,96\left(l\right)\)
c) \(NaOH+HCl\rightarrow NaCl+H_2O|\)
1 1 1 1
0,8 0,8
\(n_{NaOH}=\dfrac{0,8.1}{1}=0,8\left(mol\right)\)
\(V_{ddNaOH}=\dfrac{0,8}{2}=0,4\left(l\right)\)
Chúc bạn học tốt
\(n_{MgCO_3}=\dfrac{1,68}{84}=0,02\left(mol\right)\)
PTHH: MgCO3 + 2HCl --> MgCl2 + CO2 + H2O
______0,02--->0,04
=> \(V_{ddHCl}=\dfrac{0,04}{2}=0,02\left(l\right)\)
\(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,2\left(mol\right)\Rightarrow V_{H_2}=0,2.24,79=4,958\left(l\right)\)
b, \(n_{HCl}=2n_{Mg}=0,4\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,4.36,5}{20\%}==73\left(g\right)\)
c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,2\left(mol\right)\)
\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,2}{2,5}=0,08\left(l\right)\)
\(n_{Mg}=\dfrac{3,6}{24}=0,15\left(mol\right)\)
PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)
a, \(n_{H_2}=n_{Mg}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
b, \(n_{HCl}=2n_{Mg}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)
c, PT: \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)
\(n_{Zn}=\dfrac{9,75}{65}=0,15\left(mol\right)\)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
a, \(n_{H_2}=n_{Zn}=0,15\left(mol\right)\Rightarrow V_{H_2}=0,15.24,79=3,7185\left(l\right)\)
b, \(n_{HCl}=2n_{Zn}=0,3\left(mol\right)\Rightarrow m_{ddHCl}=\dfrac{0,3.36,5}{25\%}=43,8\left(g\right)\)
c, \(Ba\left(OH\right)_2+2HCl\rightarrow BaCl_2+2H_2O\)
Theo PT: \(n_{Ba\left(OH\right)_2}=\dfrac{1}{2}n_{HCl}=0,15\left(mol\right)\)
\(\Rightarrow V_{Ba\left(OH\right)_2}=\dfrac{0,15}{2,5}=0,06\left(l\right)\)
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: \(n_{NaOH}=n_{HCl}=0,5\left(mol\right)\) \(\Rightarrow V_{NaOH}=\dfrac{0,5}{0,5}=1\left(l\right)\)
