\(a,n_{H_2}=\dfrac{4,48}{22,4}=0,2mol\\ Zn+2HCl\rightarrow ZnCl_2+H_2\)

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

\(m_{Zn}=0,2.65=13g\\ m_{ZnO}=29,2-13=16,2g\\ b.n_{ZnO}=\dfrac{16,2}{81}=0,2mol\\ ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

0,2           0,4            0,2 

\(m_{HCl}=\left(0,4+0,4\right).36,5=29,2g\\ C_{\%HCl}=\dfrac{29,2}{200}\cdot100\%=14,6\%\\ c.m_{ZnCl_2}=\left(0,2+0,2\right).136=54,4g\) 

PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)

\(ZnO+2HCl\rightarrow ZnCl_2+H_2O\)

Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Theo PT: \(n_{Zn}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow m_{Zn}=0,2.65=13\left(g\right)\)

\(m_{ZnO}=21,1-13=8,1\left(g\right)\)

Có: \(n_{ZnO}=\dfrac{8,1}{81}=0,1\left(mol\right)\)

Theo PT: \(n_{HCl}=2n_{Zn}+2n_{ZnO}=0,6\left(mol\right)\)

\(\Rightarrow m_{HCl}=0,6.36,5=21,9\left(g\right)\Rightarrow C\%_{ddHCl}=\dfrac{21,9}{200}.100\%=10,95\%\)

Theo PT: \(n_{ZnCl_2}=n_{Zn}+n_{ZnO}=0,3\left(mol\right)\)

\(\Rightarrow m_{ZnCl_2}=0,3.136=40,8\left(g\right)\)

Bạn tham khảo nhé!

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Zn  +  2HCl →  ZnCl₂  +  H₂

ZnO  +  HCl →  ZnCl₂  +  H₂O

nH₂ = 4,48/22,4 = 0,2 mol

=> nZn = nH₂ = 0,2 mol

<=> mZn = 0,2.65= 13 gam 

<=> %mZn = \(\dfrac{13}{21,1}.100\%\) = 61,6% , %mZnO = 100 - 61,6 = 38,4%.

b. nZnO = \(\dfrac{21,1-13}{81}\) = 0,1 mol

=> nZnCl₂ = nZn + nZnO = 0,3 mol

<=> mZnCl₂ = 0,3.81 = 24,3 gam.

gọi a=nZn; b=nZnO

ta có 65a+81b= 21,1 (1)

nH2=4,48/22,4=0,2mol

Zn+2HCl->ZnCl2+H2

0,2                            0,2

ZnO+2HCl-> ZnCl2+H2O

b                    b

ta có 

nZn=a= 0,2 mol

từ (1) => b= 0,1 mol

mZn=0,2.65=13(g)

mZnO= 0,1.81=8,1g

%mZn=13.100%/21,1=61,61%

%mZnO=38,39%

mZnCl2=135.(0,2+0,1)=40,5g

\(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

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

=> \(\left\{{}\begin{matrix}\%m_{Zn}=\dfrac{0,2.65}{21,1}.100\%=61,61\%\\\%m_{ZnO}=100\%-61,61\%=38,39\%\end{matrix}\right.\)

\(n_{ZnO}=\dfrac{21,1-0,2.65}{81}=0,1\left(mol\right)\)

PTHH: ZnO + 2HCl ---> ZnCl₂ + H₂O

            0,1---->0,2------>0,1

=> \(C\%_{HCl}=\dfrac{\left(0,2+0,4\right).36,5}{200}.100\%=10,95\%\)

\(m_{mu\text{ố}i}=m_{ZnCl_2}=\left(0,1+0,2\right).136=40,8\left(g\right)\)

a)

$Mg + H_2SO_4 \to MgSO_4 + H_2$
$MgO + H_2SO_4 \to MgSO_4 + H_2O$

n Mg = n H2 = 2,24/22,4 = 0,1(mol)

%m Mg = 0,1.24/6,4  .100% = 37,5%
%m MgO = 100% -37,5% = 62,5%

b)

=> n MgO = (6,4 - 0,1.24)/40  = 0,1(mol)

=> n H2SO4 = n Mg + n MgO = 0,2(mol)

=> C% H2SO4 = 0,2.98/200  .100% = 9,8%

c)

n MgSO4 = n Mg + n MgO = 0,2(mol)

Sau phản ứng : 

m dd = 6,4 + 200 - 0,1.2 = 206,2(gam)

C% MgSO4 = 0,2.120/206,2  .100% = 11,64%

Zn+2HCl->ZnCl2+H2

ZnO+2HCl->ZnCl2+H2O

nH2=0.2(mol)->nZn=0.2(mol).mZn=13(g)

mZnO=21.1-13=8.1(g)

nZnO=0.1(mol)

Tổng nHCl cần dùng:0.2*2+0.1*2=0.6(mol)

mHCl=21.9(g)

C%ddHCl=21.9:200*100=10.95%

n muối=0.2+.1=0.3(mol)

m muối=40.8(g)

Bài 1:

a) Gọi x là số mol Zn

y là số mol ZnO

Ta có PT :

Zn + 2HCl ---> ZnCl₂ + H₂

.x.......2x..............x...........x

ZnO + 2HCl ---> ZnCl₂ + H₂O

.y...........2y..............y...........y

Theo bài ra ta có:

65x + 81y =42,2(1)

n_{\(H_2\)}=\(\frac{8,96}{22,4}\)=0,4(mol) ⇔ x=0,4

=> y = 0,2(mol)

=> m_Zn = 65.0,4=26(g)

%m_Zn = \(\frac{26}{42,2}\).100% = 61,61%

%m_ZnO = 100% - 61,61% = 38,39%

b)n_HCl = 2x + 2y = 0,4.2 + 0,2.2 = 1,2(mol)

=> m_HCl = 36,5.1,2= 43,8(g)

C%_HCl = \(\frac{43,8}{200}\).100% =21,9%

c) n_{\(ZnCl_2\)}= x + y = 0,4 + 0,2 = 0,6(mol)

=> m_{\(ZnCl_2\)} = 0,6.136 = 81,6(g)

d) Áp dụng ĐLBTKL ta có:

m_A + m_{dd HCl}= m_{dd \(ZnCl_2\)} + m_{\(H_2\)}

=>m _{dd \(ZnCl_2\)}=42,2 + 200 - 0,4.2 = 241,4(g)

C%_{\(ZnCl_2\)}= \(\frac{81,6}{241,4}\).100% = 33,8%

Bài 2

a) n_{\(H_2\)}=\(\frac{3,36}{22,4}\)=0,15(mol)

Ta có PT:

Zn + H₂SO₄ --> ZnSO₄ + H₂

0,15....0,15...........0,15....0,15

=> m_Zn = 65.0,15=9,75(g)

%m_Zn = \(\frac{9,75}{15,75}\).100%=61,9%

%m_Cu = 100% - 61,9% = 38,1%

b) m_{\(ZnSO_4\)}=161.0,15=24,15(g)