Ответ:
на твой выбор
..........................
<span>Na + Cl→ <span>Na<span>Cl
</span></span></span><span><span>3 Mg</span> + <span>2 <span>H3PO4</span></span> → <span>Mg3<span>(PO4)</span>2</span> + <span>3 <span>H2
</span></span></span><span><span>2 Al</span> + <span>3 <span>H2SO4</span></span> → <span>3 <span>H2</span></span> +<span><span>(SO4)</span><span>3
</span></span></span><span>Mg + <span>H2SO4</span> → <span>MgSO4</span> + <span>H<span>2
</span></span></span><span><span>Fe<span>(OH)</span>3</span> + <span>3 <span>HNO3</span></span> → <span>Fe<span>(NO3)</span>3</span> + <span>3 <span>H2<span>O
</span></span></span></span><span><span>Fe2O3</span> + <span>6 <span>HCl</span></span> → <span>2 <span>FeCl3</span></span> + <span>3 <span>H2<span>O
</span></span></span></span><span><span>3 <span>MgO</span></span> + <span>2 <span>H3PO4</span></span> → <span>Mg3<span>(PO4)</span>2</span> + <span>3 <span>H2<span>O
</span></span></span></span><span>Au + <span>H2SO4</span> → <span>Ø
</span></span><span><span>NaCl</span> + <span>HNO3</span> → <span>NaNO3</span> + <span>H<span>Cl</span></span></span>
<span> Дано:
W(Na2SO4)=10\% </span>Na2SO4+BaCl2=BaSO4+2NaCl <span>
m(BaCl2)=58г 1моль 1моль 1моль 2моль
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m(р-ра Na2SO4)-?
<span>Решение:
</span></span>n(BaCl2)=m/M=58г/208г/моль =0,28 моль
<span>n(Na2SO4)= 1/1 х 0,28моль =0,28 моль</span>
<span>m(Na2SO4)= n x M =0,28моль х 142г/моль=39,76 г </span>
m (p-pa Na2SO4)=m/W=39,76/0,1=397,6г<span><span>
</span> </span>
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