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

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

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

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

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

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

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

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

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

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

`a)PTHH:`

`Zn + 2HCl -> ZnCl_2 + H_2`

`0,2`   `0,4`                `0,2`      `0,2`       `(mol)`

`n_[Zn]=13/65=0,2(mol)`

`b)V_[H_2]=0,2.22,4=4,48(l)`

`c)m_[dd HCl]=[0,4.36,5]/5 . 100=292(g)`

`=>C%_[ZnCl_2]=[0,2.136]/[13+292-0,2.2].100~~8,93%`

Zn + 2HCl → ZnCl₂ + H₂

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

n Zn=0,2 mol

=>VH2=0,2.22,4=4,48l

=>mddHcl=292g

C%=\(\dfrac{0,2.136}{13+292-0,2.2}100=8,929\%\)

a. Zn (0,2 mol) + 2HCl (0,4 mol) \(\rightarrow\) ZnCl₂ (0,2 mol) + H₂ (0,2 mol).

b. Số mol Zn là 13:65=0,2 (mol).

Thể tích khí H₂ thu được ở đktc là 0,2.22,4=4,48 (l).

c. Khối lượng dung dịch là 0,4.36,5.100:5+13-0,2.2=304,6 (g).

Nồng độ phần trăm cần tìm là 0,2.136/304,6.100%\(\approx\)8,93%.

nZn=0,1 mol

Zn       +2HCl=> ZnCl2+ H2

0,1 mol =>0,2 mol

=>mHCl=36,5.0,2=7,3g

=>m dd HCl=7,3/14,6%=50g

mdd sau pứ=6,5+50-0,1.2=56,3g

=>C% dd ZnCl2=(0,1.136)/56,3.100%=24,16%

a.b.              Zn         +          2HCl        --->             ZnCl₂            +         H₂   (1)

Theo pt:     65g                     73g                            136g                        2g

Theo đề:    6,5g                   7,3g                            13,6g

=> m_ddHCl=\(\frac{7,3.100}{14,6}=50\left(g\right)\)

c. Từ pt (1), ta có: \(C_{\%}=\frac{13,6}{50+6,5}.100\%=24,1\%\)

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)

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

\(a.MgO+2HCl\rightarrow MgCl_2+H_2O\)

\(b.n_{MgO}=\dfrac{16}{40}=0,04mol\)

\(\rightarrow n_{HCl}=0,04.2=0,08mol\)

\(C_{M_{HCl}}=\dfrac{0,08}{0,15}=0,53M\)

\(c.m_{MgCl_2}=0,04.95=3,8g\)

a) $n_{CuO} = \dfrac{5}{80} = 0,0625(mol)$

$CuO + 2HCl \to CuCl_2 + H_2O$

$n_{HCl} = 2n_{CuO} = 0,125(mol)$

$m_{HCl} = 0,125.36,5 = 4,5625(gam)$

b) $m_{dd\ sau\ pư} = m_{CuO} + m_{dd\ HCl} = 5 + 500 = 505(gam)$

$C\%_{CuCl_2} = \dfrac{0,0625.135}{505}.100\% = 1,67\%$

\(CuO+2HCl\rightarrow CuCl_2+H_2O\)

\(n_{CuO}=\dfrac{5}{80}=\dfrac{1}{16}\left(mol\right)\)

Theo PTHH: \(n_{HCl}=2.n_{CuO}=\dfrac{1}{8}\left(mol\right)\)

\(m_{HCl}=\dfrac{1}{8}.36,5=\dfrac{73}{16}\left(g\right)\)

Theo PTHH: \(n_{CuCl_2}=n_{CuO}=\dfrac{1}{16}\left(mol\right)\)\(\Rightarrow m_{CuCl_2}=\dfrac{1}{16}.135=\dfrac{135}{16}\left(g\right)\)

\(C\%_{CuCl_2}=\dfrac{\dfrac{135}{16}}{500+5}.100\%\simeq1,67\%\)

a) $n_{Fe} = \dfrac{11,2}{56} = 0,2(mol)$
$Fe + 2HCl \to FeCl_2 + H_2$
$n_{HCl} =2 n_{Fe} = 0,2.2 = 0,4(mol)$
$C\%_{HCl} = \dfrac{0,4.36,5}{200}.100\% = 7,3\%$

b) $n_{H_2} = n_{FeCl_2} = n_{Fe} = 0,2(mol)

Sau phản ứng, $m_{dd} = 11,2 + 200 - 0,2.2 = 210,8(gam)$
$C\%_{FeCl_2} = \dfrac{0,2.127}{210,8}.100\% = 12,05\%$

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

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)

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