(NH4)2SO4+2NaOH=2NH3+2H2O+Na2SO4
(NH4)2SO4+BaCl2=BaSO4+ 2NH4Cl
(NH4)2SO4=2NH3+H2SO4
<span>N2→ NH3→ NH4CI→ NH3
N2+3H2=2NH3
NH3+HCl=NH4Cl
NH4Cl=</span> NH3+HCl
2KOH(изб) + H₂SO₃ = K₂SO₃ + 2H₂O
KOH + H₂SO₃(изб) = KHSO₃ + H₂O
2HCl + Na₂O = 2NaCl + H₂O
2NaOH(изб) + SO₃ = Na₂SO₄ + H₂O
NaOH + SO₃(изб) = NaHSO₄
AlCl₃ + 3AgNO₃ = 3AlCl↓ + Al(NO₃)₃
ZnCl₂ + 2KOH = Zn(OH)₂↓ + 2KCl
3NaOH + H₃PO₄ = Na₃PO₄ + 3H₂O
6HBr + Al₂O₃ = 2AlBr₃ + 3H₂O
3Ca(OH)₂ + P₂O₅ = Ca₃(PO₄)₂ + 3H₂O
Ba(NO₃)₂ + K₂SO₄ = BaSO₄↓ + 2KNO₃
CuSO₄ + 2KOH = Cu(OH)₂↓ + K₂SO₄
4NH3<span> + 3O</span>2<span> = 2N</span>2<span> + 6H</span>2<span>O
</span>N2 + O2 = 2NO
2NO + O2<span> = 2NO</span><span>2</span><span>
</span>2NO2<span> + 2NaOH (разб.) = NaNO</span>2<span> + NaNO</span>3<span> + H</span>2<span>O
</span>H3PO4<span> (конц.) + NaNO</span>3<span> = NaPO</span>3<span> + HNO</span>3<span> + H</span>2O
Дано:
m(CuSO4)=12,5гр
V(H2O)=687,5мл
найтиw(CuSO4)?
решение:
1млH2O=1грH2O
w(CuSO4)=m(ве-ва)/m(ра-ра)*100=
m(ра-ра)=687,5+12,5=700гр
w(CuSO4)=12,5/700*100=1,786%
N(Zn)=130/65=2 моль
n(Zn)/n(H2)=1/2
n(H2)=4 моль
V(H2)= 4 моль * 22,4 л/моль = 89,6 л
m(HCl)= 10\%/100\%*219г=21,9г
n(HCl) = 21,9 г/36,5г/моль=0,6 моль
n(HCl)/n(Al)=2/2
n(Al)= 0,6 моль
m(Al)=0,6 моль * 27 г/моль = 16,2 г