Не сложно догадаться, Что металлы непластичны.
Б.
C4H10 + 2Cl2 = C4H8Cl2 + 2HCl (при определенных условиях, наверное, можно получить 1,4-дихлорбутан)
СН2Cl-CH2-CH2-CH2Cl + 2Na = C4H8 + 2NaCl (C4H8 - циклобутан)
Первое:
<span><span>2</span><span>Na</span><span>O</span><span>H</span></span><span> + </span><span><span>H</span><span>2</span><span>S</span><span>O</span><span>4</span></span><span> = </span><span>Na</span><span>2</span><span>S</span><span>O</span><span>4</span><span> + </span><span>2</span><span>H</span><span>2</span><span>O</span>
<span><span>2</span><span>Na</span>+<span> + </span><span>2</span><span>O</span><span>H</span>−<span> + </span><span>2</span><span>H</span>+<span> + </span><span>S</span><span>O</span><span>4</span>2−<span> = </span><span>2</span><span>Na</span>+<span> + </span><span>S</span><span>O</span><span>4</span>2−<span> + </span><span>2</span><span>H</span><span>2</span><span>O</span></span>
<span><span><span>2</span><span>O</span><span>H</span>−<span> + </span><span>2</span><span>H</span>+<span> = </span><span>2</span><span>H</span><span>2</span><span>O</span></span></span>
<span><span><span>Второе:</span></span></span>
<span><span><span>HNO3 + K2CO3 = HK2 + CO3NO3</span></span></span>
<span><span><span><span>HNO</span>3<span> + K</span>2<span>CO</span>3<span> = HK</span>2<span> + CO</span>3<span>NO</span>3</span></span></span>
Третье:
2HCl+MgО=MgCL2+H2O
2H(+)+2Cl(-)+MgО=Mg(+2)+2CL(-)+H2O
2H(+)+MgО=Mg(+2)+H2O
Четвёртое:
<span><span>3</span><span>K</span><span>O</span><span>H</span></span><span> + </span><span><span>Fe</span>(<span>N</span><span>O</span><span>3</span>)<span>3</span></span><span> = </span><span>3</span><span>K</span><span>N</span><span>O</span><span>3</span><span> + </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span>
<span><span>3</span><span>K</span>+<span> + </span><span>3</span><span>O</span><span>H</span>−<span> + </span><span>Fe</span>3+<span> + </span><span>3</span><span>N</span><span>O</span><span>3</span>−<span> = </span><span>3</span><span>K</span>+<span> + </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span><span>+ </span><span>3</span><span>N</span><span>O</span><span>3</span>−</span>
<span><span>3</span><span>O</span><span>H</span>−<span> + </span><span>Fe</span>3+<span> = </span><span>Fe</span><span>(<span>O</span><span>H</span>)</span><span>3</span><span>+</span></span>
Пятое хз каак..
Шестое:
ZnSO4 +K3PO4=ZnSO4<span> +KOH</span>
C2H4+3O2=2CO2+2H2O
1 3
m(C2H4)=(5:22,4)*28=6,25г;m(C2H4)без примесей=6,25*0,8=5г;n(C2H4)=5г:28г/моль=0,18моль;
n(O2)=0,18моль*3=0,54моль;V(O2)=0,54моль*22,4л/моль=12л;
V(в.)=12л:0,21=57,6л
BaCO3=Ba+CO2
100%-3%=97%
m(BaCO3)=(80г*97%)/100%=77,6г
n(BaCO3)=77,6г/197г/моль=0,4 моль
n(BaCO3)=n(Ba)
m(Ba)=0,4моль*137г/моль=54,8г
Ответ:54,8г; 0,4моль