a. mdd X= 6,2+193,8= 200(g)=> C%X=\(\dfrac{6,2\cdot100}{200}=3,1\%\)

b. 2NaOH + CuSO4 ---> Na2SO4 + Cu(OH)2;

0,155--------0,0775------------0,0775----------0,0775 (mol)

Ta có: nCuSO4=\(\dfrac{200\cdot16}{100\cdot160}=0,2\left(mol\right)\)

nNaOH=\(\dfrac{6,2}{40}0,155\left(mol\right)\)

Xét tỉ lệ:\(\dfrac{nNaOH}{nNaOHpt}=\dfrac{0,155}{2}< \dfrac{nCuSO4}{nCuSO4pt}=\dfrac{0,2}{1}\)

=> CuSO4 dư. Sản phẩm tính theo NaOH.

=> nNa2SO4=0,155/2= 0,0775(mol)=> mNa2SO4=0,0775*142=11,005(g).

nCu(OH)2=0,155/2=0,0775(mol)=> mCu(OH)2=0,0775*98=7,595(g).

=> mdd sau pư= 200+200-7,595=392,405(g)

=> C%ddA=\(\dfrac{11,005\cdot100}{392,405}=2,8\%\)

c. Cu(OH)2 ---to-> CuO + H2O;

ta có: nCu(OH)2=0,0775(mol)=> nCuO=0,0775(mol)

CuO + 2 HCl --> CuCl2 + H2O;

0,0775---0,155 (mol)

nHCl=0,155(mol)=>mHCl=0,155*36,5=5,6575(g)

cho em hỏi là câu a hình như là phải tính C% NaOH chứ

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

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: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{6,2}{62}=0,2\left(mol\right)\) \(\Rightarrow C\%_{NaOH}=\dfrac{0,2\cdot40}{6,2+193,8}\cdot100\%=4\%\)

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, tính theo NaOH

\(\Rightarrow n_{Cu\left(OH\right)_2}=0,1\left(mol\right)=n_{CuO}\) \(\Rightarrow m_{CuO}=0,1\cdot80=8\left(g\right)\)

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) Na2O +H2O-->2NaOH (1)

2NaOH +CuSO4 -->Na2SO4+ Cu(OH)2 (2)

Cu(OH)2 -t^o-> CuO +H2O (3)

b) mNa2O=8.100/100=8(g)

=>nNa2O=8/62=0,13(mol)

theo(2) :nCu(OH)2=1/2nNaOH=0,065(mol)

theo(3):nCuO=nCu(OH)2=0,065(mol)

=>mCuO=0,065.80=5,2(g)

c) CuO +2HCl-->CuCl2+H2O (4)

theo (4) : nHCl=2nCuO=0,13(mol)

mddHCl 25%==18,98(g)

Na2O+H2O->2NaOH

0,1 0,1 0,2

2NaOH+CuSO4->Na2SO4+Cu(OH)2

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

a.mNaOH=0,2.40=8(g)

mdd NaOH=6,2+193,8=200(g)

C%dd NaOH=8/200.100%=4%

b.mCu(OH)2=0,1.98=9,8(g)

c.Cu(OH)2->CuO+H2O

0,1 0,1 0,1

CuO+2HCl->CuCl2+H2O

0,1 0,2

VddHCl=0,2/2=0,1(l)

Câu c mình ko biết làm đúng hay ko

Bài này em làm đúng rồi nhé