\(n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ a,Na_2O+H_2O\rightarrow2NaOH\\ NaOH+HCl\rightarrow NaCl+H_2O\\ n_{HCl}=n_{NaCl}=0,2.2=0,4mol\\ b,C_{M_{HCl}}=\dfrac{0,4}{0,2}=2M\\ c,m_{ddNaCl}=200.1,05+12,4=222,4g\\ C_{\%NaCl}=\dfrac{0,4.58,5}{222,4}\cdot100=10,52\%\)

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

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

\(n_{Zn}=\dfrac{19.5}{65}=0.3\left(mol\right)\)

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

\(0.3........0.6.........0.3......0.3\)

\(m_{HCl}=0.3\cdot36.5=10.95\left(g\right)\)

\(C\%_{HCl}=\dfrac{10.95}{200}\cdot100\%=5.475\%\)

\(m_{\text{dung dịch sau phản ứng}}=19.5+200-0.3\cdot2=218.9\left(g\right)\)

\(m_{ZnCl_2}=0.3\cdot136=40.8\left(g\right)\)

\(C\%_{ZnCl_2}=\dfrac{40.8}{218.9}\cdot100\%=18.63\%\)

mình đang cần gấp

;-;tui cũng đang thắc mắc câu này

Theo đề bài ta có : nH2 = 10,08/22,4 = 0,45 (mol)

a) PTHH :

0,45mol->,9mol->0,45mol

b) khối lượng mạt sắt tham gia phản ứng là :

mFe = 0,45.56 = 25,2(g)

c)

nồng độ mol của dd HCl đã dùng là :

CMddHCl = 0,9/0,15 = 6(M)

nCO2=0,4(mol)

a) PTHH: 2 NaOH + CO2 -> Na2CO3 + H2O

0,8_________0,4________0,4(mol)

=> mNaOH=0,8.40=32(g) 

=>C%ddNaOH=(32/200).100=16%

b) mddNa2CO3=mddNaOH+mCO2=200+0,4.44=217,6(g)

mNa2CO3=106.0,4=42,4(g)

=>C%ddNa2CO3=(42,4/217,6).100=19,485%

Chúc em học tốt!

n_CO2=8,96/22,4=0,4mol

a/ CO₂+2NaOH→Na₂CO₃+H₂O

   0,4      0,8           0,4         0,4

m_NaOH=0,8.40=32g

C%dd_NaOH=m_ct/_mdd.100%=32/200.100%=16%

b/m_CO2=0,4.44=17,6g

Theo định luật bảo toàn khối lượng:

m_CO2+m_NaOH=m_Na2CO3

17,6g+200g=217,6g

m_Na2CO3=0,4.106=42,4g

C%ddNa2CO3=m_ct/mdd.100%=42,4/217,6.100=19,4852g

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

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

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

\(n_{CuSO4}=\dfrac{16\%.50}{100\%.160}=0,05\left(mol\right)\)

Pt : \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

       0,05-------->0,1---------->0,05--------->0,05

a) \(C\%_{ddNaOH}=\dfrac{0,1.40}{250}.100\%=1,6\%\)

b) \(m_{ddspu}=50+250-0,05.98=295,1\left(g\right)\)

\(C\%_{Na2SO4}=\dfrac{0,05.142}{295,1}.100\%=2,41\%\)

\(n_{CuSO_4}=\dfrac{m_{dd}\cdot C\%}{100\cdot M}=\dfrac{50\cdot16\%}{100\cdot\left(64+32+16\cdot4\right)}=0,05\left(mol\right)\)

\(PTHH:CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)

                     1                  2                    1                    1

                   0,05              0,1                 0,05               0,05 (mol)

\(a)C\%_{NaOH}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{01\cdot100\cdot\left(23+16+1\right)}{250}=1,6\%\)

\(b)m_{dd-sau-pư}=m_{dd_đ}+m_{ct_đ}-m\downarrow-m\uparrow\)

\(=50+250-\left(0,05\cdot23+32+16\cdot4\right)=294,05\left(g\right)\)

\(C\%_{Na_2SO_4}=\dfrac{n\cdot100\cdot M}{m_{dd}}=\dfrac{0,05\cdot100\cdot\left(23\cdot2+32+16\cdot4\right)}{294,05}\approx2,41\%.\)

nCuSO4​=100%.16016%.50​=0,05(mol)

Pt : ����4+2����→��(��)2+��2��4CuSO4​+2NaOH→Cu(OH)2​+Na2​SO4​

       0,05-------->0,1---------->0,05--------->0,05

a) �%������=0,1.40250.100%=1,6%C%ddNaOH​=2500,1.40​.100%=1,6%

b) ������=50+250−0,05.98=295,1(�)mddspu​=50+250−0,05.98=295,1(g)

�%��2��4=0,05.142295,1.100%=2,41%C%Na2SO4​=295,10,05.142​.100%=2,41%