<span><span>4Fe</span>+<span>3<span>O2 </span></span>→ <span>2<span>Fe2O3
</span></span></span><span><span>Fe2O3</span>+<span>6<span>HCl</span></span> → <span>2<span>FeCl3</span></span>+<span>3<span>H2O </span></span></span>
<span><span>FeCl3</span>+<span>3<span>NaOH</span></span> → <span>3<span>NaCl</span></span>+<span>Fe<span>(OH)</span>3
</span></span><span><span>2<span>Fe<span>(OH)</span>3</span></span>+<span>3<span>H2SO4</span></span> → <span>Fe2<span>(SO4)</span>3</span>+<span>6<span>H2<span>O
</span></span></span></span>
2)О2 и Н3РО4
я не виноват если неправильно
Правельный ответ 4), так как Fe(|||)
Ответ: 2
CO2 + NaOH = NaHCO3
HCl + NaOH = NaCl + H2O
3) m(NH3)=V(NH3)/Vm*M(NH3)=11,2/22,4*17=0,03(г)
объем раствора 500 + 0,03 = 500,03 г
массовая доля = 0,03/500,03 *100% = 0,6% аммиака