<span>ну, например так</span>
<span>CuSO</span><span>₄</span> +2<span>KOH</span> = <span>Cu</span>(<span>OH</span>)<span>₂↓</span> + <span>K</span><span>₂</span><span>SO</span><span>₄ - молекулярное</span>
<span>Cu</span><span>²⁺ + </span><span>SO</span><span>₄²⁻ + 2</span><span>K</span><span>⁺ + 2</span><span>OH</span><span>⁻ = </span><span>Cu</span><span>(</span><span>OH</span><span>)₂↓ + 2</span><span>K</span><span>⁺ +</span><span>SO</span><span>₄²⁻ - полное ионное</span>
<span>Cu</span><span>²⁺ +</span><span> 2</span><span>OH</span><span>⁻</span> = <span>Cu</span><span>(</span><span>OH</span><span>)₂↓ - сокращенное ионное</span>
<span> </span>
Оксид хлора С|2О, кислород О2, вода Н2О, углекислый газ СО2, Оксид меди CuO
2KOH + CuSO4 → K2SO4 + Cu(OH)2 ↓
m(KOH) = 22.5 г → n ≈ 0.4
m(CuSO4) = 12.5 г → n ≈ 0.08 убыток
0.08 • 98 = 7.84 г
7.84 – 100%
икс - 85%
ответ: икс = (85•7.84)/100=6.664 г (Cu(OH)2)
AlCl3+4NaOH=Na[Al(OH)4]+3NaCl