nHCl = 0,017 mol

nNaOH = 0,004 mol

NaOH (0,004) + HCl_dư (0,004) ------> NaCl + H2O (1)

- Theo PTHH(1): = > nHCl dư = 0,004 mol

=> nHCl phản ứng = 0,017 - 0,004 = 0,013 mol

Gọi nZnO là x thì nMgO là y

=> 81x + 40y = 0,3 (I)

ZnO (x) + 2HCl (2x) -------> ZnCl2 + H2O (2)

MgO (y) + 2HCl (2y) -------> MgCl2 + H2O (3)

-Theo PTHH(1,2) => nHCl = 2x +2y = 0,013 (II)

- Giải hệ PT(I,II) => \(\left\{{}\begin{matrix}x=\dfrac{1}{1025}\left(mol\right)\\y=\dfrac{453}{82000}\left(mol\right)\end{matrix}\right.\)

=> mZnO = \(\dfrac{1}{1025}.81=\dfrac{81}{1025}\left(gam\right)\)

=> mMgO = \(\dfrac{453}{82000}.40=\dfrac{453}{2050}\left(gam\right)\)

PT: \(ZnO+2HCl\rightarrow ZnCl_2+H_2O\) (1)

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

\(HCl+NaOH\rightarrow NaCl+H_2O\) (3)

Số mol NaOH cần dùng để trung hòa HCl dư là:

\(n_{NaOH}=0,008.0,5=0,004\left(mol\right)\)

Theo pt (3): \(n_{NaOH}=n_{HCl}=0,004\left(mol\right)\)

Hay: \(n_{HCl_{dư}}=0,004\left(mol\right)\)

Số mol HCl tham gia pư là :

\(n_{HCl}=0,017.1=0,017\left(mol\right)\)

\(\Rightarrow n_{HCl_{1+2}}=0,017-0,004=0,013\left(mol\right)\)

Giả sử: Số mol ZnO là x(mol)

Số mol MgO là y (mol)

(x,y>0)

\(\Rightarrow m_{ZnO}=81x\left(g\right)\)

\(m_{MgO}=40y\left(g\right)\)

\(\Rightarrow81x+40y=0,3\) (*)

Theo pt (1) : \(n_{HCl_1}=2.n_{ZnO}=2x\left(mol\right)\)

(2) :\(n_{HCl_2}=2.n_{MgO}=2y\left(mol\right)\)

Mà: \(n_{HCl_{1+2}}=0,013\left(mol\right)\)

\(\Rightarrow2x+2y=0,013\) (**)

Từ (*) và (**),ta có hệ phương trình:

\(\left\{{}\begin{matrix}81x+40y=0,3\\2x+2y=0,013\end{matrix}\right.\)

Giải hệ phương trình,ta đc:

\(\left\{{}\begin{matrix}x=\dfrac{1}{1025}\left(mol\right)=n_{ZnO}\\y=\dfrac{453}{82000}\left(mol\right)=n_{MgO}\end{matrix}\right.\)

\(\Rightarrow m_{ZnO}=\dfrac{1}{1025}.81=\dfrac{81}{1025}\left(g\right)\)

\(m_{MgO}=\dfrac{453}{82000}.40=\dfrac{453}{2050}\left(g\right)\)

=.= hok tốt!!

$NaOH + HCl \to NaCl + H_2O$
$n_{HCl\ dư} = n_{NaOH} = 0,05.2 = 0,1(mol)$

Gọi $n_{Fe} = a ; n_{Zn} = b \Rightarrow 56a + 65b = 12,1(1)$

$Fe + 2HCl  \to FeCl_2 + H_2$
$Zn + 2HCl \to ZnCl_2 + H_2$
$n_{HCl} = 2a + 2b = 0,5  -0,1 = 0,4(2)$

Từ (1)(2) suy ra  a = b = 0,1

$\%m_{Fe} = \dfrac{0,1.56}{12,1}.100\% = 46,28\%$

$\%m_{Zn} = 100\% -46,28\% = 53,72\%$

\(Fe+2HCl\rightarrow FeCl_2+H_2\)

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

\(HCl_{dư}+NaOH\rightarrow NaCl+H_2O\)

Gọi x, y lần lượt là số mol Fe, Zn, theo đề ta có:

\(\left\{{}\begin{matrix}56x+65y=12,1\\2x+2y=0,5.1-0,05.2\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=0,1\\y=0,1\end{matrix}\right.\)

=> \(\%m_{Fe}=\dfrac{0,1.56}{12,1}=46,28\%\)

=> \(\%m_{Zn}=100-46,28=53,72\%\)

Đáp án D

\(n_{H_2SO_4}=0,4.0,5=0,2\left(mol\right)\)

=> \(n_{H_2O}=0,2\left(mol\right)\)

Theo ĐLBTKL: m_oxit + m_H2SO4 = m_muối + m_H2O

=> m_muối = 8 + 0,2.98 - 0,2.18 = 24 (g)

n _H2SO4=0,5.0,4=0,2 mol

=> m  muối =8+0,2.96=27,2g

\(HCOOH+NaOH\rightarrow HCOONa+H_2O\)

\(CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O\)

\(n_{HCOOH}=a\left(mol\right),n_{CH_3COOH}=b\left(mol\right)\)

\(m_X=46a+60b=10.6\left(g\right)\left(1\right)\)

\(n_{NaOH}=a+b=0.2\left(mol\right)\left(2\right)\)

\(\Rightarrow a=b=0.1\)

\(m_{HCOOH}=0.1\cdot46=4.6\left(g\right)\)

\(m_{CH_3COOH}=6\left(g\right)\)

\(n_{H_2O}=0.2\left(mol\right)\)

\(BTKL:\)

\(m_{Muối}=10.6+0.2\cdot40-0.2\cdot18=15\left(g\right)\)

 Dung` DL BTKL: moxit + mH2SO4 = mmuoi' + mH2O 
voi' nH2O = nH2SO4 = 0.5*0.1 = 0.05 
--> mmuoi' = 2.81 + 0.05*98 - 0.05*18 = 6.81g 

Cach' #: (Fe2O3, MgO, ZnO) ----> (Fe2(SO4)3; MgSO4, ZnSO4) 
--> nO = nSO4(2-) = nH2SO4 = 0.05 
--> m(Fe, Mg, Zn) = 2.81 - mO = 2.81 - 0.05*16 = 2.01g 

mmuoi' = mKL + mSO4(2-) = 2.01 + 0.05*96 = 6.81g 

500 ml = 0,5 l

Ta có n_H2SO4 = 0,5 . 0,1 = 0,05 (mol)

Fe2O3 + 3H2SO4 \(\rightarrow\) Fe2(SO4)3 + 3H2O (1)

x \(\rightarrow\) 3x \(\rightarrow\) x \(\rightarrow\) 3x

H2SO4 + MgO \(\rightarrow\) MgSO4 + H2O (2)

y \(\rightarrow\) y \(\rightarrow\) y \(\rightarrow\) y

H2SO4 + ZnO \(\rightarrow\) ZnSO4 + H2O (3)

z \(\rightarrow\) z \(\rightarrow\) z \(\rightarrow\) z

từ các phương trình (1)(2)(3) => n_H2 = n_H2SO4 = 0,05

Theo định luật bảo toàn khối lượng

=> M_muối = M_{kim loại} + M_axít​ - M_H2

= 2,81 + 0,05 . 98 - 0,05 . 2

= 7,61 ( gam )

m muối = m oxit + 80*n h2so4
= 2.81 + 80*0.05= 6.81g
đơn giản dễ hiểu

$a\big)$

$Zn+2CH_3COOH\to (CH_3COO)_2Zn+H_2$

$ZnO+2CH_3COOH\to (CH_2COO)_2Zn+H_2O$

Theo PT: $n_{Zn}=n_{H_2}=\frac{4,48}{22,4}=0,2(mol)$

$\to \%m_{Zn}=\frac{0,2.65}{21,1}.100\%\approx 61,61\%$

$\to \%m_{ZnO}=100-61,61=38,39\%$

$b\big)$

$n_{ZnO}=\frac{21,1-0,2.65}{81}=0,1(mol)$

Theo PT: $\sum n_{CH_3COOH}=2n_{Zn}+2n_{ZnO}=0,6(mol)$

$\to C_{M_{CH_3COOH}}=\dfrac{0,6}{\frac{200}{1000}}=3M$

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

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

 0,2                                                                0,2     ( mol )

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

\(\%m_{Zn}=\dfrac{13}{21,1}.100=61,61\%\)

\(\%m_{ZnO}=100\%-61,61\%=38,39\%\)

\(Zn+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2\)

0,2               0,4                                                  ( mol )

\(n_{ZnO}=\dfrac{21,1-13}{81}=0,1mol\)

\(ZnO+2CH_3COOH\rightarrow\left(CH_3COO\right)_2Zn+H_2O\)

 0,1              0,2                                                        ( mol )

\(C_{M\left(CH_3COOH\right)}=\dfrac{0,4+0,2}{0,2}=3M\)

Chọn A