\(n_{Zn}=\dfrac{13}{65}=0,2(mol)\\ a,PTHH:Zn+2HCl\to ZnCl_2+H_2\\ b,n_{HCl}=0,4(mol)\\ \Rightarrow C\%_{HCl}=\dfrac{0,4.36,5}{100}.100\%=14,6\%\\ c,n_{ZnCl_2}=n_{H_2}=0,2(mol)\\ \Rightarrow m_{ZnCl_2}=0,2.136=27,2(g)\\ \Rightarrow C\%_{ZnCl_2}=\dfrac{27,2}{13+100-0,2.2}.100\%\approx 24,16\%\)

Ủa thịnh THCS HTA đúng kh? Này đề thi giữa kì hóa mà thịnh đi hỏi hả ;) tui méc thầy nha

Chúc cậu học tốt nhee

a, \(Mg+2HCl\rightarrow MgCl_2+H_2\)

b, Ta có: \(n_{Mg}=\dfrac{2,4}{24}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Mg}=0,2\left(mol\right)\Rightarrow C\%_{HCl}=\dfrac{0,2.36,5}{400}.100\%=1,825\%\)

c, Theo PT: \(n_{MgCl_2}=n_{H_2}=n_{Mg}=0,1\left(mol\right)\)

Ta có: m _{dd sau pư} = 2,4 + 400 - 0,1.2 = 402,2 (g)

\(\Rightarrow C\%_{MgCl_2}=\dfrac{0,1.95}{402,2}.100\%\approx2,36\%\)

\(n_{Al2O3}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

Pt : \(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O|\)

          1             6              2            3

         0,1         0,6            0,1 

a) \(n_{HCl}=\dfrac{0,1.6}{1}=0,6\left(mol\right)\)

\(m_{HCl}=0,6.36,5=21,9\left(g\right)\)

\(m_{ddHCl}=\dfrac{21,9.100}{10}=219\left(g\right)\)

b) \(n_{AlCl3}=\dfrac{0,6.2}{6}=0,2\left(mol\right)\)

⇒ \(m_{AlCl3}=0,2.133,5=26,7\left(g\right)\)

c) \(m_{ddspu}=10,2+219=229,2\left(g\right)\)

\(C_{AlCl3}=\dfrac{26,7.100}{229,2}=11,65\)⁰/₀ 

 Chúc bạn học tốt

nAl2O3=10.2:102=0.1(mol) 

PTHH:Al2O3+6HCl->2AlCl3+3H2O

theo pthh:nHCl:nAl2O3=6->nHCl=6*0.1=0.6(mol)

mHCl=0.6*36.5=21.9(g)

mdd HCl=21.9*100:14.6=150(g)

theo pthh:nAlCl3:nAl2O3=2->nAlCl3=0.1*2=0.2(mol)

mAlCl3=0.2*133.5=26.7(g)

mdd sau phản ứng:10.2+150=160.2

a)\(n_{Al_2O_3}=\dfrac{m}{M}=\dfrac{10,2}{102}=0,1\left(mol\right)\)

PTHH:\(Al_2O_3+6HCl\rightarrow2AlCl_3+3H_2O\)

             0,1          0,6           0,2         0,3         (mol)

\(m_{HCl}=n.M=0,6.36,5=21,9\left(g\right)\)

b)\(m_{AlCl_3}=n.M=0,2.133,5=26,7\left(g\right)\)

c)\(m_{ddHCl}=\dfrac{m_{HCl}}{C\%}.100\)=\(\dfrac{21,9}{10}.100\)=219(g)

theo định luật bảo toàn khối lượng

\(m_{Al_2O_3}+m_{ddHCl}=m_{ddAlCl_3}\)\(\Rightarrow m_{ddAlCl_3}=10,2+219\)=229,2(g)

\(C\%_{AlCl_3}=\dfrac{m_{AlCl_3}}{m_{ddAlCl_3}}.100\)=\(\dfrac{26,7}{229,2}.100\)=8,92%

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\)

a) Pt : \(Zn+2HCl\rightarrow ZnCl_2+H_2|\)

           1          2               1        1

         0,2       0,4            0,2       0,2

b) \(n_{HCl}=\dfrac{0,2.2}{1}=0,4\left(mol\right)\)

⇒ \(m_{HCl}=0,4.36,5=14,6\left(g\right)\)

\(C_{ddHCl}=\dfrac{14,6.100}{100}=14,6\)⁰/₀

c) \(n_{ZnCl2}=\dfrac{0,4.1}{2}=0,2\left(mol\right)\)

⇒ \(m_{ZnCl2}=0,2.136=27,2\left(g\right)\)

\(m_{ddspu}=13+100-\left(0,2.2\right)=112,6\left(g\right)\)

\(C_{ZnCl2}=\dfrac{27,2.100}{112,6}=24,16\)⁰/₀

 Chúc bạn học tốt

\(n_{Zn}=\dfrac{13}{65}=0.2\left(mol\right)\)

\(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(0.2........0.4..........0.2.......0.2\)

\(m_{HCl}=0.4\cdot36.5=14.6\left(g\right)\)

\(C\%_{HCl}=\dfrac{14.6}{100}\cdot100\%=14.6\%\)

\(m_{ZnCl_2}=0.2\cdot136=27.2\left(g\right)\)

\(m_{\text{dung dịch sau phản ứng}}=13+100-0.2\cdot2=112.6\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{27.2}{112.6}\cdot100\%=24.1\%\)

PTHH: \(2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\)

Ta có: \(n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{1}{2}n_{NaOH}=\dfrac{1}{2}\cdot\dfrac{10}{40}=0,125\left(mol\right)\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddH_2SO_4}=\dfrac{0,125\cdot98}{10\%}=122,5\left(g\right)\\m_{Na_2SO_4}=0,125\cdot142=17,75\left(g\right)\end{matrix}\right.\)

\(\Rightarrow C\%_{Na_2SO_4}=\dfrac{17,75}{10+122,5}\cdot100\%\approx13,4\%\)

\(n_{Zn}=\dfrac{13}{65}=0,2\left(mol\right)\\ pthh:Zn+2HCl\rightarrow ZnCl_2+H_2\) 
          0,2        0,4                      0,2 
\(V_{H_2}=0,2.22,4=4,48\left(l\right)\\ C\%_{HCl}=\dfrac{0,4.36,5}{200}.100\%=7,3\%\)

a) 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

b) \(n_{NaOH}=\dfrac{100.20}{100.40}=0,5\left(mol\right)\)

PTHH: 2NaOH + H₂SO₄ -- Na₂SO₄ + 2H₂O

______0,5----->0,25------>0,25

=> m_H2SO4 = 0,25.98 = 24,5 (g)

=> \(m_{ddH_2SO_4}=\dfrac{24,5.100}{19,6}=125\left(g\right)\)

c) m_Na2SO4 = 0,25.142 = 35,5 (g)

m_{dd sau pư} = 100 + 125 = 225 (g)

=> \(C\%\left(Na_2SO_4\right)=\dfrac{35,5}{225}.100\%=15,778\%\)

Ta có: \(m_{NaOH}=200.20\%=40\left(g\right)\Rightarrow n_{NaOH}=\dfrac{40}{40}=1\left(mol\right)\)

PT: \(NaOH+HCl\rightarrow NaCl+H_2O\)

a, Theo PT: \(n_{HCl}=n_{NaOH}=1\left(mol\right)\)

\(\Rightarrow C\%_{HCl}=\dfrac{0,1.36,5}{100}.100\%=36,5\%\)

b, \(n_{NaCl}=n_{NaOH}=1\left(mol\right)\)

Ta có: m _{dd sau pư} = 200 + 100 = 300 (g)

\(\Rightarrow C\%_{NaCl}=\dfrac{1.58,5}{300}.100\%=19,5\%\)

a) \(n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2\left(mol\right)\)

PTHH: Na₂CO₃ + 2HCl --> 2NaCl + CO₂ + H₂O

                0,2------------------>0,4---->0,2

m_{dd sau pư} = 200 + 21,2 - 0,2.44 = 212,4(g)

=> \(C\%\left(NaCl\right)=\dfrac{0,4.58,5}{212,4}.100\%=11,017\%\)

$PTHH:Na_2CO_3+2HCl\to 2NaCl+H_2O+CO_2\uparrow$

$n_{Na_2CO_3}=\dfrac{21,2}{106}=0,2(mol)$

Theo PT: $n_{NaCl}=n_{CO_2}=0,2(mol)$

$\Rightarrow m_{NaCl}=0,4.58,5=23,4(g);m_{CO_2}=0,2.44=8,8(g)$

$\Rightarrow C\%_{NaCl}=\dfrac{23,4}{21,2+200-8,8}.100\%\approx 11,01\%$