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

Ta có: \(n_{CuO}=\dfrac{4}{80}=0,05\left(mol\right)\)

\(\Rightarrow\left\{{}\begin{matrix}n_{HCl}=0,1\left(mol\right)\\n_{CuCl_2}=0,05\left(mol\right)\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m_{ddHCl}=\dfrac{0,1\cdot36,5}{7,3\%}=50\left(g\right)\\C\%_{CuCl_2}=\dfrac{0,05\cdot135}{4+50}\cdot100\%=12,5\%\end{matrix}\right.\)

\(n_{MgO}=\dfrac{4}{40}=0,1\left(mol\right)\\ PTHH:MgO+2HCl\rightarrow MgCl_2+H_2O\\ \Rightarrow n_{HCl}=2n_{MgO}=0,2\left(mol\right)\\ \Rightarrow m_{CT_{HCl}}=0,2\cdot36,5=7,3\left(g\right)\\ \Rightarrow m_{dd_{HCl}}=\dfrac{7,3\cdot100\%}{7,3\%}=100\left(g\right)\\ n_{MgCl_2}=n_{H_2}=0,1\left(mol\right)\\ \Rightarrow\left\{{}\begin{matrix}m_{CT_{MgCl_2}}=0,1\cdot95=9,5\left(g\right)\\m_{H_2}=0,1\cdot18=1,8\left(g\right)\end{matrix}\right.\\ \Rightarrow m_{dd_{MgCl_2}}=4+100-1,8=102,2\left(g\right)\\ \Rightarrow C\%_{MgCl_2}=\dfrac{9,5}{102,2}\cdot100\%\approx9,3\%\)

\(n_{MgO}=\dfrac{4}{40}=0,1mol\)

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

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

\(m_{HCl}=0,2\cdot36,5=7,3\)

\(m_{ddHCl}=\dfrac{7,3}{7,3}\cdot100=100g\)

\(m_{MgCl_2}=0,1\cdot95=9,5g\)

\(m_{H_2O}=0,1\cdot18=1,8g\)

\(m_{ddsaupu}=4+100-1,8=102,2g\)

\(\%m_{m'}=\dfrac{9,5}{102,2}\cdot100\%=9,3\%\)

\(n_{CuO}=\dfrac{8}{80}=0,1\left(mol\right)\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)

             1            2             1            1

             0,1        0,2           0,1

b) \(n_{HCl}=\dfrac{0,1.2}{1}=0,2\left(mol\right)\)

⇒ \(m_{HCl}=0,2.36,5=7,3\left(g\right)\)

\(C_{ddHCl}=\dfrac{7,3.100}{200}=3,65\)⁰/₀

c) \(n_{CuCl2}=\dfrac{0,2.1}{2}=0,1\left(mol\right)\)

⇒ \(m_{CuCl2}=0,1.135=13,5\left(g\right)\)

 Chúc bạn học tốt

a)m_H2SO4=\(\dfrac{200.7,3\text{%}}{100\%}\)=14,6g

n_HCl=\(\dfrac{14,6}{36,5}\)=0,4(mol)

PTHH:

NaOH+ HCl→ NaCl+ H₂O

1            1            1         1

0,4        0,4         0,4     (mol)

⇒m_NaOH=0,4.40=16(g)

Nồng độ % của dd NaOH cần dùng là:

C_%NaOH=\(\dfrac{16}{200}\)   .100%=8%

b)Ta có:m_{dd spứ}=m_{dd trc pứ}=400g

m_NaCl=0,4.58,5=23,4g

Nồng độ % dd muối tạo thành sau pứ là:

C_%dd NaCl=\(\dfrac{23,4}{400}\)   .100%=5,85%        

) Gọi n Na2CO3= x ; n Na2So3 =y

=> 106x + 126y= 55 (I)

Ta có: mHCl =36,5 => nHCl= 1

Na2CO3 + 2HCl --> 2NaCl + CO2 + H2O (1)

x 2x 2x x

Na2SO3 + 2HCl --> 2NaCl + SO2 + H2O (2)

y 2y 2y y

=> 2x + 2y = 1 (II)

(I) và (II) => n Na2CO3= 0,4 và n Na2SO3 = 0,1

(1)=> nCO2 = 0,4

(2) => nSO2= 0,1

=> %V CO2 = 80% => %V SO2 = 20%

b) ΣΣ nNaCl= 1=> mNaCl= 58,5

m dd sau pứ= 55 + 250 - 0,4 x 44 - 0,1 x 64= 281(g)

C% dd NaCl = 20,82%

a)

$AgNO_3 + HCl \to AgCl + HNO_3$
Theo PTHH : 

$n_{AgCl} = n_{HCl} = n_{AgNO_3} = \dfrac{340.10\%}{170} =0,2(mol)$

$m_{dd\ HCl} = \dfrac{0,2.36,5}{7,3\%} = 100(gam)$

b)

$m_{AgCl} = 0,2.143,5 = 28,7(gam)$

c)

$m_{dd\ sau\ pư} = 340 + 100 -28,7 = 411,3(gam)$

$n_{HNO_3} = n_{AgNO_3} = 0,2(mol)$
$\Rightarrow C\%_{HNO_3} = \dfrac{0,2.63}{411,3}.100\% = 3,06\%$

a) $n_{CaCO_3} = 0,15(mol)$
$CaCO_3 + 2HCl \to CaCl_2 + CO_2 + H_2O$
$n_{HCl} = 2n_{CaCO_3} = 0,3(mol)$
$m_{dd\ HCl} = \dfrac{0,3.36,5}{7,3\%} = 150(gam)$

b)

$n_{CaCl_2} = n_{CO_2} = n_{CaCO_3} =0,15(mol)$
$V_{CO_2} = 0,15.22,4 = 3,36(lít)$
c)

$m_{dd} = 15 + 150 - 0,15.44 = 158,4(gam)$

$C\%_{CaCl_2} = \dfrac{0,15.111}{158,4}.100\% = 10,51\%$

\(nAl=\dfrac{2,7}{27}=0,1\left(mol\right)\)

\(mHCl=\dfrac{200.7,3\%}{100\%}=14,6\left(g\right)\)

\(nHCl=\dfrac{14,6}{36,5}=0,4\left(mol\right)\)

PTHH:

\(2Al+6HCl\rightarrow2AlCl_2+3H_2\)

 2         6              2            3    (mol)

0,1       0,3         0,1           0,15   (mol)

LTL : 0,1 / 2 < 0,4/6

=> Al đủ , HCl dư

1. \(VH_2=0,15.22,4=3,36\left(l\right)\)

2. \(mH_2=0,15.2=0,3\left(g\right)\)

mdd = mAl + mddHCl - mH2 = 2,7 + 200 - 0,3 = 202,4 (g)

\(mH_2SO_{4\left(dưsaupứ\right)}=0,1.98=9,8\left(g\right)\)

\(mAlCl_2=0,1.98=9,8\left(g\right)\)

\(C\%_{ddH_2SO_4}=\dfrac{9,8.100}{202,4}=4,84\%\)

\(C\%_{AlCl_2}=\dfrac{9,8.100}{202,4}=4,84\%\)