Bài 2:
\(a,n_{H_2}=\dfrac{1,12}{22,4}=0,05(mol)\\ PTHH:Mg+2HCl\to MgCl_2+H_2\\ MgO+2HCl\to MgCl_2+H_2O\\ \Rightarrow n_{Mg}=0,05(mol)\\ \Rightarrow m_{Mg}=24.0,05=1,2(g)\\ \Rightarrow m_{MgO}=9,2-1,2=8(g) b,\%_{Mg}=\dfrac{1,2}{9,2}.100\%=13,04\%\\ \Rightarrow \%_{MgO}=100\%-13,04\%=86,96\%\\ c,n_{MgO}=\dfrac{8}{40}=0,2(mol)\\ \Rightarrow \Sigma n_{HCl}=2n_{Mg}+2n_{MgO}=0,5(mol)\\ \Rightarrow \Sigma m_{HCl}=0,5.36,5=18,25(g)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{18,25}{14,6\%}=125(g)\)
Bài 5 :
\(n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
Pt : \(Mg+2HCl\rightarrow MgCl_2+H_2|\)
1 2 1 1
0,05 0,1 0,05
\(MgO+2HCl\rightarrow MgCl_2+H_2O|\)
1 2 1 1
0,2 0,4
a) \(n_{Mg}=\dfrac{0,05.1}{1}=0,05\left(mol\right)\)
\(m_{Mg}=0,05.24=1,2\left(g\right)\)
\(m_{MgO}=9,2-1,2=8\left(g\right)\)
b) 0/0Mg = \(\dfrac{1,2.100}{9,2}=13,04\)0/0
0/0MgO = \(\dfrac{8.100}{9,2}=86,96\)0/0
c) Có : \(m_{MgO}=8\left(g\right)\)
\(n_{MgO}=\dfrac{8}{40}=0,2\left(mol\right)\)
\(n_{HCl\left(tổng\right)}=0,1+0,4=0,5\left(mol\right)\)
\(m_{HCl}=0,5.36,5=18,25\left(g\right)\)
\(m_{ddHCl}=\dfrac{18,25.100}{14,6}=125\left(g\right)\)
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Bài 3:
\(a,PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow\\ b,n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)\\ \Rightarrow n_{CO_2}=0,2(mol)\\ \Rightarrow V_{CO_2}=0,2.22,4=4,48(l)\\ c,m_{dd_{HCl}}=1,15.300=345(g)\\ n_{NaCl}=0,4(mol);n_{H_2O}=n_{CO_2}=0,2(mol)\\ \Rightarrow m_{NaCl}=0,4.58,5=23,4(g)\\ m_{H_2O}=18.0,2=3,6(g)\\ m_{CO_2}=44.0,2=8,8(g)\\ m_{dd_{NaCl}}=21,2+345-3,6-8,8=353,8(g)\\ \Rightarrow C\%_{NaCl}=\dfrac{23,4}{353,8}.100\%=6,61\%\)
B3:
\(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\\ a,Na_2CO_3+2HCl\rightarrow2NaCl+CO_2+H_2O\\ b,n_{CO_2}=n_{Na_2CO_3}=0,2\left(mol\right)\\ V_{CO_2\left(\text{đ}ktc\right)}=0,2.22,4=4,48\left(l\right)\\ c,m_{\text{dd}sau}=m_{Na_2CO_3}+m_{\text{dd}HCl}-m_{CO_2}=21,2+300.1,15-0,2.44=357,4\left(g\right)\\ n_{NaCl}=2.0,2=0,4\left(mol\right)\\ C\%_{\text{dd}NaCl}=\dfrac{0,4.58,5}{357,4}.100\approx3,274\%\)
