thấy khi nằm đúng thì tick nha

nNH3=0.05 mol

nCuO=0.2 mol

3CuO+2NH3->3Cu+N2+3H2O

0.075 0.05

nCuO dư=0.125

CuO+2HCl->CuCl2+H2

0.125 0.25

->VHCl=0.25/0.5=0.5L

nNH3=0.4 mol

4NH3+502->4NO+6H2O

0.4 0.5

->VO2=11,2 l

nMg=0.05 mol

Mg+2HCl->MgCl2+H2

0.05 0,1

mHCl=3.65g->mdd HCl=25g

AO+2HCl->ACl2+H2O

0.05 0.1

->MAO=80G/MOL

->MA=64g/mol->A là Cu

2)nCuO=0.2 mol

CuO+2HCl->CuCl2+H2O

nHCl=0.4 mol

mdd HCl=200 g

m muối=27g

->%dd muối=\(\dfrac{27}{16+200}\)*100=12,5%

B

Đặt công thức trung bình của 2kl là A

2A+2HCl->2ACl+H2

m muối=m kim loại+m Cl

->mCl=19,15-8,5=10.65g

->nCl=0.3 mol

mà nCl=nH->nH=0.3 mol->nH2=0,15 mol->VH2=3,36 l

22,4 mà bạn

Fe+2HCl->FeCl2+H2

0.1 0.1(theo PTHH)

->VH2=2.24l

D:a,c,d

\(\Leftrightarrow\sqrt[4]{3x+1}-\sqrt[4]{x+1}+\sqrt[4]{4x+1}-\sqrt[4]{x+1}=0\)

\(\Leftrightarrow\frac{3x+1-\left(x+1\right)}{A^3+A^2B+AB^2+B^3}+\frac{4x+1-\left(x+1\right)}{C^3+C^2B+CB^2+B^3}=0\)

\(\left(A=\sqrt[4]{3x+1};\text{ }B=\sqrt[4]{x+1};\text{ }C=\sqrt[4]{4x+1};\text{ }A,B>0;\text{ }C\ge0\right)\)

\(\Leftrightarrow x\left(\frac{2}{....}+\frac{3}{.....}\right)=0\)

\(\Leftrightarrow x=0.\)

NaOH+HCl->NaCl+H2O

x mol x mol

KOH+HCl->KCl+H2O

y mol y mol

Theo bài ra:40x+56y=3.04

58,5x+74,5y=4.15

->x=0.02 mol->mNaOH=0.8g
y=0.04 mol->mKOH=2.24g