PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

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

            \(Cu\left(OH\right)_2\xrightarrow[]{t^o}CuO+H_2O\)

a) Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{193,8+6,2}\cdot100\%=4\%\)

b) Ta có: \(\left\{{}\begin{matrix}n_{NaOH}=0,2\left(mol\right)\\n_{CuSO_4}=\dfrac{200\cdot16\%}{160}=0,2\left(mol\right)\end{matrix}\right.\)

Xét tỉ lệ: \(\dfrac{0,2}{2}< \dfrac{0,2}{1}\) \(\Rightarrow\) CuSO₄ còn dư, NaOH p/ứ hết

\(\Rightarrow n_{Cu\left(OH\right)_2}=\dfrac{1}{2}n_{NaOH}=n_{CuO}=0,1\left(mol\right)\)

\(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)

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

Theo PTHH: \(n_{HCl}=2n_{CuO}=0,2\left(mol\right)\) \(\Rightarrow V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(l\right)=100\left(ml\right)\)

a)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5}{5+45}\cdot100\%=10\%\)

b)\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{5,6}{5,6+94,4}\cdot100\%=5,6\%\)

c)\(m_{ctNaOH}=\dfrac{200\cdot10\%}{100\%}=20g\)

   \(m_{ctNaOH}=\dfrac{300\cdot5\%}{100\%}=15g\)

\(C\%=\dfrac{m_{ct}}{m_{dd}}\cdot100\%=\dfrac{20+15}{200+300}\cdot100\%=7\%\)

\(a,C\%_{NaOH}=\dfrac{5}{5+45}=10\%\)

b, \(n_{CaO}=\dfrac{5,6}{56}=0,1\left(mol\right)\)

PTHH: CaO + H₂O ---> Ca(OH)₂

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

\(\rightarrow C\%_{Ca\left(OH\right)_2}=\dfrac{74.0,1}{5,6+94,4}=37\%\)

c, \(m_{NaOH}=10\%.200+5\%.300=35\left(g\right)\)

\(\rightarrow C\%_{NaOH}=\dfrac{35}{200+300}=7\%\)

Na2O=0,5 mol

Na2O+H2O->2NaOH

0,5-----------------1 mol

ta có m NaOH=1.40+40=80g

=>C%=\(\dfrac{80}{431}100=18,561\%\)

a. Khối lượng chất tan: 10g

khối lượng dung dịch : 100g

b.

Khối lượng chất tan: 6,2 g

khối lượng dung dịch :111,6g

c.

Khối lượng chất tan: 4,6g

khối lượng dung dịch :154,6g

d.

Khối lượng chất tan: 6,5g

khối lượng dung dịch :206,5g

\(a,m_{ct}=10\left(g\right)\\ m_{dd}=10+90=100\left(g\right)\)

b, \(n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\)

PTHH: Na₂O + H₂O ---> 2NaOH

            0,1                          0,2

\(m_{ct}=0,2.40=8\left(g\right)\\ m_{dd}=6,2+105,4=111,6\left(g\right)\)

c, \(n_{Na}=\dfrac{4,6}{23}=0,2\left(mol\right)\)

PTHH: 2Na + 2H₂O ---> 2NaOH + H₂

            0,2                        0,2        0,1

m_ct = 0,2.40 = 8 (g)

\(m_{dd}=4,6+150-0,1.2=154,4\left(g\right)\)

d, \(n_{Zn}=\dfrac{6,5}{65}=0,1\left(mol\right)\)

PTHH: Zn + 2HCl ---> ZnCl₂ + H₂

            0,1                     0,1      0,1

m_ct = 0,1.136 = 13,6 (g)

m_dd = 200 + 6,5 - 0,2 = 206,3 (g)

Câu 1 : 

\(n_{Al}=\dfrac{2.7}{27}=0.1\left(mol\right)\)

\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)

\(0.1......................0.1.........0.15\)

\(m_{dd}=2.7+200-0.15\cdot2=202.4\left(g\right)\)

\(C\%_{AlCl_3}=\dfrac{0.1\cdot133.5}{202.4}\cdot100\%=6.59\%\)

a, \(Na_2O+2HCl\rightarrow2NaCl+H_2O\)

b, Ta có: \(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

Theo PT: \(n_{NaCl}=2n_{Na_2O}=0,4\left(mol\right)\)

\(\Rightarrow C\%_{NaCl}=\dfrac{0,4.58,5}{12,4+200}.100\%\approx11,02\%\)

\(n_{Na_2O}=\dfrac{12,4}{62}=0,2\left(mol\right)\)

a.

\(Na_2O+2HCl\rightarrow2NaCl+H_2O\)

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

b.

\(C\%_{NaCl}=\dfrac{0,4.58,5.100\%}{12,4+200,}=11,02\%\)

`HaNa♬D`

\(a/Na_2O+2HCl\rightarrow2NaCl+H_2O\\ b/n_{Na_2O}=\dfrac{12,4}{62}=0,2mol\\ n_{NaCl}=2.0,2=0,4mol\\ C_{\%NaCl}=\dfrac{0,4.58,5}{200+12,4}\cdot100=11,02\%\)

\(m_{Ct\left(NaOH\right)}=200\cdot2\%=4g\)

=>\(n_{NaOH}=\dfrac{4}{40}=0.1\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

0,05                      0,1                     

\(m_{Na_2O}=0.05\cdot\left(23\cdot2+16\right)=3.1\left(g\right)\)

\(n_{CO_2}=\dfrac{V}{22,4}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\\ n_{NaOH}=\dfrac{m}{M}=\dfrac{m_{dd}.C\%}{M}=\dfrac{200.16\%}{40}=0,8\left(mol\right)\)

Có: \(\dfrac{n_{NaOH}}{n_{CO_2}}=4\)

=> Phản ứng tạo muối Na₂CO₃

PT:

\(CO_2+2NaOH\rightarrow Na_2CO_3+H_2O\)

0,2.       0,4            0,2

=> dd sau phản ứng có những chất tan là:

\(\left\{{}\begin{matrix}Na_2CO_3:0,2\left(mol\right)\\NaOH:0,4\left(mol\right)\end{matrix}\right.\)

m_dd spu=0,2.44+200=208,8(g)

\(\%m_{NaOH}=\dfrac{0,4.40}{208,8}.100\%=7,66\%\\\%m_{Na_2CO_3}=\dfrac{0,2.106}{208,8}.100\%=10,15\% \)

Câu 1:

nAl= 0,1(mol)

PTHH: 2Al + 6 HCl -> 2 AlCl3 + 3 H2 

nAlCl3= nAl=0,1(mol)

-> mAlCl3= 133,5 x 0,1= 13,35(g)

mddAlCl3= mAl + mddHCl - mH2 = 2,7 + 200 - 3/2 x 0,1 x 2= 202,4(g)

C%ddAlCl3= (13,35/202,4).100= 6,596%

bạn ơi, có nồng độ của dd HCl không? vì nếu làm ntn thì câu 1 đề bài không chặt chẽ

\(a.n_{Na_2O}=\dfrac{6,2}{62}=0,1\left(mol\right)\\ Na_2O+H_2O\rightarrow NaOH\\ m_{ddNaOH}=193,8+6,2=200\left(g\right)\\C\%_{ddX}=C\%_{ddNaOH}=\dfrac{0,1.2.40}{200}.100=4\%\\ b.2NaOH+CuSO_4\rightarrow Cu\left(OH\right)_2+Na_2SO_4\\ a=m_{Cu\left(OH\right)_2}=\dfrac{0,2}{2}.98=9,8\left(g\right)\\ c.Cu\left(OH\right)_2\underrightarrow{to}CuO+H_2O\\ CuO+2HCl\rightarrow CuCl_2+H_2O\\ n_{HCl}=2.n_{CuO}=2.n_{Cu\left(OH\right)_2}=2.0,1=0,2\left(mol\right)\\ V_{ddHCl}=\dfrac{0,2}{2}=0,1\left(lít\right)=100\left(ml\right)\)