<span>№1.
AICI</span>₃⇄AI⁺³ + CI⁻
NaOH ⇄Na⁺ + OH⁻
H₂S₂O₃⇄2H⁺ + S₂O₃²⁻
CuSO₄⇄ Cu²⁺ + SO₄²⁻
H₂SO₃⇄2H⁺ + SO₃²⁻
<span>№2
CuCI</span>₂+2NaOH=Cu(OH)₂↓ + 2NaCI
Cu²⁺ + 2CI⁻+2Na⁺ + 2OH⁻= Cu(OH)₂↓ + 2Na⁺ + 2CI⁻
Cu²⁺ + 2OH⁻= Cu(OH)₂↓<span>
3Mg(NO</span>₃)₂+2Na₃PO₄=Mg₃(PO₄)₂↓ + 6NaNO₃
3Mg²⁺ + 6NO₃⁻+6Na⁺ + 2PO₄³⁻= Mg₃(PO₄)₂↓ + 6Na⁺ + 6NO₃⁻
3Mg²⁺ + 2PO₄³⁻= Mg₃(PO₄)₂↓
<span>
Fe</span>₂O₃+3H₂SO₄=Fe₂(SO₄)₃ + 3H₂O
Fe₂O₃+6H⁺ + 3SO₄²⁻ = 2Fe³⁺ +3SO₄ + 3H₂O
Fe₂O₃+6H⁺ = 2Fe³⁺ + 3H₂O
<span>№3.
Zn</span>²⁺+ CO₃²⁻→ZnCO₃<span>
ZnCI</span>₂ + Na₂CO₃ = ZnCO₃ + 2NaCI
<span>
CuO +2H</span>⁺ →Cu²⁺ +H₂O
CuO + 2HCI = CuCI₂ + H₂O
.........................................
w вещ-ва = m вещ-ва / m р-ра * 100 %
m р-ра = m вещ-ва + m(H2O)
m р-ра = 200 г + 20 г = 220 г
w вещ-ва (сахара) = 20 / 220 * 100 % = 9,1 %
Ответ: 9,1 %