Задание 7 ЕГЭ по химии

3415

А) Zn — набор 3: $H_{2}O,\ HCl,\ CuSO_{4}$

• $Zn + H_{2}O \overset{t}{\rightarrow} ZnO + H_{2}\uparrow$

• $Zn + 2HCl = ZnCl_{2} + H_{2}\uparrow$

• $Zn + CuSO_{4} = ZnSO_{4} + Cu\downarrow$

Б) CuO — набор 4: $H_{2}SO_{4},\ H_{2},\ CO$

• $CuO + H_{2}SO_{4} = CuSO_{4} + H_{2}O$

• $CuO + H_{2} \overset{t}{\rightarrow} Cu + H_{2}O$

• $CuO + CO \overset{t}{\rightarrow} Cu + CO_{2}$

В) $Ca(OH)_{2}$ — набор 1: $SiO_{2},\ K_{3}PO_{4},\ Br_{2}$

• $Ca(OH)_{2} + SiO_{2} \overset{t}{\rightarrow} CaSiO_{3} + H_{2}O$

• $3Ca(OH)_{2} + 2K_{3}PO_{4} = Ca_{3}(PO_{4})_{2}\downarrow + 6KOH$

• $2Ca(OH)_{2} + 2Br_{2} = Ca(BrO)_{2} + CaBr_{2} + 2H_{2}O$

Г) $K_{2}SO_{4}$ — набор 5: $BaCl_{2},\ (CH_{3}COO)_{2}Ba,\ H_{2}SO_{4}$

• $K_{2}SO_{4} + BaCl_{2} = BaSO_{4}\downarrow + 2KCl$

• $K_{2}SO_{4} + (CH_{3}COO)_{2}Ba = BaSO_{4}\downarrow + 2CH_{3}COOK$

• $K_{2}SO_{4} + H_{2}SO_{4} = 2KHSO_{4}$

Набор 2 не подходит ни одному веществу: $Zn + Cu \cancel{=}$, $CuO + Cu \cancel{=}$, $Ca(OH)_{2} + Cu \cancel{=}$, $K_{2}SO_{4} + NaOH \cancel{=}$.

Ответ: 3415