Дано:
![m _{p} (NaOH)=](https://tex.z-dn.net/?f=m+_%7Bp%7D+%28NaOH%29%3D)
200 г
![w(NaOH)=](https://tex.z-dn.net/?f=w%28NaOH%29%3D)
5%=0,05
![m(H _{2} SO _{4} )=](https://tex.z-dn.net/?f=m%28H+_%7B2%7D+SO+_%7B4%7D+%29%3D)
29,4 г
![w(H _{2} SO _{4} )=](https://tex.z-dn.net/?f=w%28H+_%7B2%7D+SO+_%7B4%7D+%29%3D)
50%=0,5
Решение:
1) составляем уравнение :
![2NaOH+H _{2} SO _{4} =Na _{2} SO _{4} +2H_{2} O](https://tex.z-dn.net/?f=2NaOH%2BH+_%7B2%7D+SO+_%7B4%7D+%3DNa+_%7B2%7D+SO+_%7B4%7D+%2B2H_%7B2%7D+O)
2) находим массу
![NaOH](https://tex.z-dn.net/?f=NaOH)
![\\](https://tex.z-dn.net/?f=+%5C%5C+)
![m(NaOH)=m _{p} (NaOH)*w(NaOH)=](https://tex.z-dn.net/?f=m%28NaOH%29%3Dm+_%7Bp%7D+%28NaOH%29%2Aw%28NaOH%29%3D)
=
![200*0.5=100](https://tex.z-dn.net/?f=200%2A0.5%3D100)
г
3) находим химическое количество
![NaOH](https://tex.z-dn.net/?f=NaOH)
![\\](https://tex.z-dn.net/?f=+%5C%5C+)
![n(NaOH)= \frac{m(NaOH)}{M(NaOH)} = \frac{100}{40} =2.5](https://tex.z-dn.net/?f=n%28NaOH%29%3D+%5Cfrac%7Bm%28NaOH%29%7D%7BM%28NaOH%29%7D+%3D+%5Cfrac%7B100%7D%7B40%7D+%3D2.5)
моль
4) находим массу
![m(H _{2} SO _{4} )=m _{p} (H _{2} SO _{4} )*w(H _{2} SO _{4} )=29.4*0.5=14.7](https://tex.z-dn.net/?f=m%28H+_%7B2%7D+SO+_%7B4%7D+%29%3Dm+_%7Bp%7D+%28H+_%7B2%7D+SO+_%7B4%7D+%29%2Aw%28H+_%7B2%7D+SO+_%7B4%7D+%29%3D29.4%2A0.5%3D14.7)
г
5) находим химическое количество
![H _{2} SO _{4}](https://tex.z-dn.net/?f=H+_%7B2%7D+SO+_%7B4%7D+)
![\\](https://tex.z-dn.net/?f=+%5C%5C+)
![n(H _{2} SO_{4} )= \frac{n(H _{2} SO _{4} )}{M(H _{2} SO _{4} )} = \frac{14.7}{98} =0.15](https://tex.z-dn.net/?f=n%28H+++_%7B2%7D+++SO_%7B4%7D+%29%3D+%5Cfrac%7Bn%28H+_%7B2%7D+SO+_%7B4%7D+%29%7D%7BM%28H+_%7B2%7D+SO+_%7B4%7D+%29%7D+%3D+%5Cfrac%7B14.7%7D%7B98%7D+%3D0.15)
моль
6)
2,5 моль -
![n(NaOH)](https://tex.z-dn.net/?f=n%28NaOH%29)
0,15 моль -
![n(H _{2} SO_{4} )](https://tex.z-dn.net/?f=n%28H+_%7B2%7D+SO_%7B4%7D+%29)
мы получаем , что взят
![NaOH](https://tex.z-dn.net/?f=NaOH)
в избытке, а в недостатке
значит дальше вычисления проводим по
7) находим химическое количество
![n(Na _{2} SO_{4} )=n(H _{2} SO _{4} )=0.15](https://tex.z-dn.net/?f=n%28Na+_%7B2%7D+SO_%7B4%7D+%29%3Dn%28H+_%7B2%7D+SO+_%7B4%7D+%29%3D0.15)
моль
8) находим массу
![m(Na_{2} SO _{4})](https://tex.z-dn.net/?f=m%28Na_%7B2%7D+SO+_%7B4%7D%29+)
![\\](https://tex.z-dn.net/?f=+%5C%5C+)
г
![m(Na_{2} SO _{4})=n(Na _{2} SO _{4} )*M(Na _{2} SO _{4} )=0.15*142=21.3](https://tex.z-dn.net/?f=m%28Na_%7B2%7D+SO+_%7B4%7D%29%3Dn%28Na+_%7B2%7D+SO+_%7B4%7D+%29%2AM%28Na+_%7B2%7D+SO+_%7B4%7D+%29%3D0.15%2A142%3D21.3)
г
Ответ:
![m(Na_{2} SO _{4}) =21.3](https://tex.z-dn.net/?f=m%28Na_%7B2%7D+SO+_%7B4%7D%29+%3D21.3)
г
Cu3(PO4)2 это соль . Ортофосфат меди
1.<span>NH2CH2COOH+NaOH->NH2CH2COONa+H2O</span>
2Ba + O2 -t> 2BaO
BaO + H2O -> Ba(OH)2
Ba(OH)2 + 2HCl - > BaCl2 + 2H2O