..........................................
<span><span>(NH<span>4)2SO4 + BaCl2 ---> 2NH4Cl + BaSO4 </span></span>
<span><span>2NH<span>4(+) + SO4(2-) + Ba(2+) + 2Cl(-) ---> 2NH4(+) + 2Cl(-) + BaSO4 </span></span>
<span><span>SO<span>4(2-) + Ba(2+) ---> BaSO4 </span></span></span></span></span>
Решение представлено во вложении
1) NaCl---> Na(+)+Cl(-)
2)<span>KNO3---></span><span>K(+)+NO3(-)</span>
<span>3)<span>CaCO3---></span><span>не диссоц</span></span>
<span><span>4)<span>Na3PO4--->3</span><span>Na(+)+PO4(3-)</span></span></span>
<span><span><span>5)<span>Cu(NO3)2---><span>Cu(2+)+2NO3(-)</span></span></span></span></span>
<span><span><span><span><span>6)<span>BaSO4--- не диссоц</span></span></span></span></span></span>
<span><span><span><span><span><span>7)<span>Fe(NO3)2---><span>Fe(2+)+ 2NO3(-)</span></span></span></span></span></span></span></span>
<span><span><span><span><span><span><span><span>8)<span>Fe2(SO4)3--->2</span><span>Fe(3+)+3SO4(2-)</span></span></span></span></span></span></span></span></span>
<span><span><span><span><span><span><span><span><span>9)<span>Ca3(PO4)2--- не диссоц</span></span></span></span></span></span></span></span></span></span>
<span><span><span><span><span><span><span><span><span><span>10) <span>K2SiO3--->2</span><span>K(+)+ SiO3(2-)</span></span></span></span></span></span></span></span></span></span></span>
W(S)=2M(S)/M(FeS2)=64/120*100\%=53,3\%