nCO2=0.025(mol)

nCa(OH)2=0.02(mol)

\(\frac{nCa\left(OH\right)2}{nCO2}=\frac{0.02}{0.025}=0.8\) 

->tạo cả muối trung hòa và muối axit

CO2+Ca(OH)2->CaCO3+H2O (1)

2CO2+Ca(OH)2->Ca(HCO3)2 (2)

Gọi nCO2(1) là x(mol)->nCa(OH)2=x(mol)

nCO2(2) là y(mol)->nCa(OH)2=1/2 y(mol)

theo bài ra ta có:x+y=0.025

x+1/2 y=0.02

x=0.015(mol)->nCaCO3=0.015(mol)->mCaCO3=1.5(g)

y=0.01(mol)->nCa(HCO3)2=0.05(mol)->mCa(HCO3)2=8.1(g)

m muối=8.1+1.5=9.6(g)

nCa(OH)2=0.5(mol)

Ca(OH)2+CO2->CaCO3+H2O(1)

Ca(OH)2+2CO2->Ca(HCO3)2(2)

nCaCO3=0.35(mol)

theo pthh nCO2(1)=nCa(OH)2=nCaCO3=0.35(mol)

->nCO2(1)=nCa(OH)2=0.35(mol)

nCa(OH)2(2)=0.5-0.35=0.15(mol)

nCO2=3nCa(OH)2=0.3(mol)

tổng nCO2=0.3+0.35=0.65(mol)

V=14.56(l)

nCa(HCO3)2=0.15(mol)

CM muối tan=0.15:2=0.075(M)

Fe+2HCl-> FeCl2+H2

nH2=3.36:22.4=0.15(mol)

Theo pthh nFe=nH2

-> nFe phản ứng=0.15(mol)

mFe=0.15*56=8.4(g)

nHCl=2 nH2->nHCl phản ứng=0.3(mol)

nHCl dư=1.4-0.3=1.1(mol)

mHCl dư=1.1*36.5=40.15(g)

nhầm

1002 chia hết cho 3

1008 chia hết cho 9

nCaCl2=0.03(mol)

nAgNO3=0.02(mol)

CaCl2+2AgNO3->Ca(NO3)2+2AgCl

Theo pthh nAgNO3=2nCaCl2

Theo bài ra,nCaCl2=1.5nCaCl2

->CaCl2 dư tính theo AgNO3

nAgCl=0.02(mol)

mAgCl=2.87(g)

nCa(NO3)=0.01(mol)

CM=0.01:(0.03+0.07)=0.1

nCaCl2 dư=0.3-0.2:2=0.2(mol)

CM=0.2:(0.03+0.07)=2(M)

nSO2=0.25(mol)

Cu+2H2SO4->CuSo4+SO2+2H2O

CuO+H2SO4->CuSO4+H2O

nCu=nSO2=0.25(mol)

mCu=16(g)

->mCuO=12(g)

nCuO=0.15(mol)

mH2SO4=78.4

nH2SO4=0.8(mol)

tổng nH2SO4 phản ứng:0.5+0.15=0.65(mol)

nH2SO4 dư=0.15(mol)

mH2SO4 dư=14.7(g)

nCuSO4=0.4(mol)

mCuSO4=64(g)

mdd=28+112-64*0.25=124(g)

C%(H2SO4)=14.7:124*100=11.9%

C%(CuSO4)=64:124*100=51.6%

BÀI 2

mdd axit=900(g)

mH2SO4=220.5(g)

gọi mSO2 là x(g)

ta có m chất tan sau khi hòa tan=x+220.5

mdd sau khi hòa tan=x+900

theo bài ra:(x+220.5):(x+900)=49/100

100x+22050=49x+44100

51x=22050

->x=432.4(g)

nAl2O3=0.05(mol)

Al2O3+3H2SO4->Al2(SO4)3+3H2O

Theo pthh nH2SO4=3 nAl2O3->nH2SO4=0.15(mol)

mH2SO4=14.7(g)

mDd H2SO4=14.7*100:4.7=313(g)

nAl2(SO4)3=0.05(mol)

mAl2(SO4)3=17.1(g)

mdd sau phản ứng:318.1(g)

C% muối=17.1:318.1*100=5.4%