\(n_{Mg}=\dfrac{4.8}{24}=0.2\left(mol\right)\)

\(Mg+2HCl\rightarrow MgCl_2+H_2\)

\(0.2.....................0.2..........0.2\)

\(V_{H_2}=0.2\cdot22.4=4.48\left(l\right)\)

\(m_{\text{dung dịch sau phản ứng }}=4.8+250-0.2\cdot2=254.4\left(g\right)\)

\(m_{MgCl_2}=0.2\cdot95=19\left(g\right)\)

\(C\%_{MgCl_2}=\dfrac{19}{254.4}\cdot100\%=7.47\%\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

a, PT: \(Mg+2HCl\rightarrow MgCl_2+H_2\)

____0,2____0,4______0,2____0,2 (mol)

b, \(V_{H_2}=0,2.22,4=4,48\left(l\right)\)

c, Ta có: m _{dd sau pư} = m_Mg + m _{dd HCl} - m_H2 = 4,8 + 250 - 0,2.2 = 254,4 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,2.95}{254,4}.100\%\approx7,47\%\)

Bạn tham khảo nhé!

Theo gt ta có: $n_{Mg}=0,2(mol)$

a, $Mg+2HCl\rightarrow MgCl_2+H_2$

b, Ta có: $n_{H_2}=0,2(mol)\Rightarrow V_{H_2}=4,48(l)$

c, Ta có: $m_{dd}=4,8+250-0,2.2=254,4(g)$

$\Rightarrow \%C_{MgCl_2}=7,47\%$

Bài 1: 

PTHH: \(Mg+2HCl\rightarrow MgCl_2+H_2\uparrow\)

Ta có: \(n_{Mg}=\dfrac{4,8}{24}=0,2\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,4\left(mol\right)\\n_{H_2}=0,2\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,4\cdot36,5}{14,6\%}=100\left(g\right)\\V_{H_2}=0,2\cdot22,4=4,48\left(l\right)\end{matrix}\right.\)

Bài 2:

PTHH: \(2KOH+H_2SO_4\rightarrow K_2SO_4+2H_2O\)

Ta có: \(\left\{{}\begin{matrix}n_{KOH}=\dfrac{100\cdot11,2\%}{56}=0,2\left(mol\right)\\n_{H_2SO_4}=\dfrac{150\cdot9,8\%}{98}=0,15\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,15}{1}\) \(\Rightarrow\) H₂SO₄ còn dư, KOH p/ứ hết

\(\Rightarrow\left\{{}\begin{matrix}n_{K_2SO_4}=0,1\left(mol\right)\\n_{H_2SO_4\left(dư\right)}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{K_2SO_4}=0,1\cdot174=17,4\left(g\right)\\m_{H_2SO_4\left(dư\right)}=0,05\cdot98=4,9\left(g\right)\end{matrix}\right.\)

Mặt khác: \(m_{dd}=m_{ddKOH}+m_{ddH_2SO_4}=250\left(g\right)\)

\(\Rightarrow\left\{{}\begin{matrix}C\%_{K_2SO_4}=\dfrac{17,4}{250}\cdot100\%=6,96\%\\C\%_{H_2SO_4\left(dư\right)}=\dfrac{4,9}{250}\cdot100\%=1,96\%\end{matrix}\right.\)

\(a)n_{H_2}=\dfrac{7,437}{24,79}=0,3mol\\ 2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2\)

0,2        0,3               0,1                  0,3

\(\%m_{Al}=\dfrac{0,2.27}{15,6}\cdot100=34,62\%\\ \%m_{Al_2O_3}=100-34,62=65,38\%\\ b)n_{Al_2O_3}=\dfrac{\left(15,6-0,2.27\right)}{102}=0,1mol\)

\(2Al+3H_2SO_4\rightarrow Al_2\left(SO_4\right)_3+3H_2O\)

0,1        0,15           0,05                 0,15

\(m_{ddH_2SO_4}=\dfrac{\left(0,3+0,15\right).98}{9,8}\cdot100=450g\)

c) \(C_{\%Al_2\left(SO_4\right)_3}=\dfrac{\left(0,1+0,05\right)342}{15,6+450-0,3.2}\cdot100=11\%\)

Ta có: \(n_{H_2}=\dfrac{13,44}{22,4}=0,6\left(mol\right)\)

BTNT, có: \(n_{SO_4}=n_{H_2SO_4}=n_{H_2}=0,6\left(mol\right)\)

Mà: m _muối = m_KL + m_SO4

⇒ m = m_KL = 93,6 - 0,6.96 = 36 (g)

Bạn tham khảo nhé!

\(n_{H_2}=\dfrac{1.12}{22.4}=0.05\left(mol\right)\)

\(Zn+H_2SO_4\rightarrow ZnSO_4+H_2\)

\(0.05.......0.05......0.05...........0.05\)

\(m_{Zn}=0.05\cdot65=3.25\left(g\right)\)

\(C_{M_{H_2SO_4}}=\dfrac{0.05}{0.2}=0.25\left(M\right)\)

\(m_{ZnSO_4}=0.05\cdot161=8.05\left(g\right)\)

\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)

\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)

\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)

a) 2Al + 6HCl --> 2AlCl₃ + 3H₂

Fe + 2HCl --> FeCl₂ + H₂

b) Gọi số mol Al, Fe lần lượt là a,b 

=> 27a + 56b = 13,9

\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)

2Al + 6HCl --> 2AlCl₃ + 3H₂

a----->3a--------->a------->1,5a______(mol)

Fe + 2HCl --> FeCl₂ + H₂

b------>2b-------->b----->b__________(mol)

=> 1,5a + b = 0,35

=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)

c) n_HCl = 3a + 2b = 0,7 (mol)

=> m_HCl = 0,7.36,5 = 25,55(g)

=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)

\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)

\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)

B1:

2NaOH+H₂SO₄\(\rightarrow\)Na₂SO₄+2H₂O

n_NaOH=\(\frac{4}{40}=0.1\)mol

=>n_H2SO4=\(\frac{1}{2}\)n_NaOH=0.05 mol

=>C_M=\(\frac{n_{H2SO42}}{V}\)=\(\frac{0.05}{200}\)=2,5.10^-4 (M)

B2:

Mg+\(\frac{1}{2}\)O₂\(\underrightarrow{t^0}\)MgO                (1)

MgO+2HCl\(\rightarrow\)MgCl₂+H₂O (2)

n_Mg(1)=\(\frac{0,36}{24}=0,015mol\)

=>n_MgO(1)=0,015=n_MgO(2)

n_HCl(2)=2n_MgO(2)=0,03mol

=>C_M(HCl)=\(\frac{n_{HCl}}{V}=\frac{0,03}{100}=3.10^{-4}M\)

$n_{HCl} = 0,8.0,5 = 0,4(mol) ; n_{H_2SO_4} = 0,6(mol) ;n_{H_2} = 0,2(mol)$

$n_{H(trong\ axit)} = 0,4 + 0,6.2 = 1,6(mol)$

Bảo toàn H : $n_{H_2O} = \dfrac{n_{H(trong\ axit)} - 2n_{H_2} }{2} = 0,6(mol)$
Bảo toàn khối lượng : 

$m = 88,7 + 0,6.18 + 0,2.2 - 0,4.36,5 -  0,6.98 = 26,5(gam)$

Tính nồng độ phần trăm của dd Z nhé ạ.       

Gọi \(n_{H_2}=5a\left(mol\right)\) \(\Rightarrow n_{CO_2}=11a\left(mol\right)\)

\(\Rightarrow5a+11a=\dfrac{3,584}{22,4}\) \(\Rightarrow a=0,01\) \(\Rightarrow\left\{{}\begin{matrix}n_{H_2}=0,05\left(mol\right)\\n_{CO_2}=0,11\left(mol\right)\end{matrix}\right.\) \(\Rightarrow m_{khí}=0,05\cdot2+0,11\cdot44=4,94\left(g\right)\)

Ta có: \(n_{HCl}=\dfrac{188\cdot1,25\cdot7,3\%}{36,5}=0,47\left(mol\right)\) \(\Rightarrow m_{HCl}=0,47\cdot36,5=17,155\left(g\right)\) 

Bảo toàn Hidro: \(n_{H_2O}=\dfrac{1}{2}n_{HCl}=0,235\left(mol\right)\) \(\Rightarrow m_{H_2O}=0,235\cdot18=4,23\left(g\right)\)

Bảo toàn khối lượng: \(m_{NaCl}=m_{hhX}+m_{HCl}-m_{khí}-m_{H_2O}=27,945\left(g\right)\)

Mặt khác: \(m_{ddHCl}=188\cdot1,25=235\left(g\right)\)

\(\Rightarrow m_{dd\left(sau.pư\right)}=m_{hhX}+m_{ddHCl}-m_{khí}=250,02\left(g\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{27,945}{250,02}\cdot100\%\approx11,18\%\)