Задание 7 ЕГЭ по химии

3541

А) Cu → 3

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

• $Cu + 2AgNO_{3} = Cu(NO_{3})_{2} + 2Ag\downarrow$

• $Cu + 4HNO_{3(конц.)} = Cu(NO_{3})_{2} + 2NO_{2}\uparrow + 2H_{2}O$

Б) $Fe_{2}O_{3}$ → 5

• $2Fe_{2}O_{3} + 3C \overset{t}{\rightarrow} 4Fe + 3CO_{2}\uparrow$

• $Fe_{2}O_{3} + 3CO \overset{t}{\rightarrow} 2Fe + 3CO_{2}\uparrow$

• $Fe_{2}O_{3} + 6HBr = 2FeBr_{3} + 3H_{2}O$

В) $H_{2}SO_{4}$ (р-р) → 4

• $ZnO + H_{2}SO_{4} = ZnSO_{4} + H_{2}O$

• $2NaOH + H_{2}SO_{4} = Na_{2}SO_{4} + 2H_{2}O$

• $Na_{2}SiO_{3} + H_{2}SO_{4} = Na_{2}SO_{4} + H_{2}SiO_{3}\downarrow$

Г) $NaHCO_{3}$ → 1

• $AlCl_{3} + 3NaHCO_{3} = Al(OH)_{3}\downarrow + 3NaCl + 3CO_{2}\uparrow$

• $NaHCO_{3} + NaOH = Na_{2}CO_{3} + H_{2}O$

• $NaHCO_{3} + CH_{3}COOH = CH_{3}COONa + H_{2}O + CO_{2}\uparrow$

Набор 2 лишний — в нём есть реакции, которые не идут: $Cu + H_{2}O \cancel{=}$, $Fe_{2}O_{3} + KCl \cancel{=}$, $H_{2}SO_{4(р-р)} + KCl \cancel{=}$, $NaHCO_{3} + Mg(NO_{3})_{2} \cancel{=}$.

Ответ: 3541