<span>1)196х0.75:98х160:0.8^2=375<span>
</span></span>
Ответ:
дано
m(NaCL) = 200 g
------------------
m(Na)-?
M(NaCL) = 58.5 g/mol
M(Na) = 23 g/mol
200 g NaCL --------- 58.5 g/mol NaCL
Xg Na ----------------- 23 g/mol Na
X = 200 * 23 / 58.5 = 78.6 g
ответ 78.6 г
Объяснение:
![[tex]3CaCl _{2} +2Na _{3} PO _{4} =Ca _{3}( PO _{4} ) _{2}+6NaCl \\ \\ m(CaCl _{2} )=m*w=300*0.111=33.3 \\n(CaCl _{2} )= \frac{33.3}{111} =0.3\\\\m(Na _{3} PO _{4})=m*w=200*0.246=49.2\\n(Na _{3} PO _{4})= \frac{49.2}{23*3+31+64} = \frac{49.2}{164} =0.3 \\ \\ \\](https://tex.z-dn.net/?f=%5Btex%5D3CaCl%20_%7B2%7D%20%2B2Na%20_%7B3%7D%20PO%20_%7B4%7D%20%3DCa%20_%7B3%7D%28%20PO%20_%7B4%7D%20%29%20_%7B2%7D%2B6NaCl%20%5C%5C%20%5C%5C%20m%28CaCl%20_%7B2%7D%20%29%3Dm%2Aw%3D300%2A0.111%3D33.3%20%5C%5Cn%28CaCl%20_%7B2%7D%20%29%3D%20%5Cfrac%7B33.3%7D%7B111%7D%20%3D0.3%5C%5C%5C%5Cm%28Na%20_%7B3%7D%20PO%20_%7B4%7D%29%3Dm%2Aw%3D200%2A0.246%3D49.2%5C%5Cn%28Na%20_%7B3%7D%20PO%20_%7B4%7D%29%3D%20%5Cfrac%7B49.2%7D%7B23%2A3%2B31%2B64%7D%20%3D%20%5Cfrac%7B49.2%7D%7B164%7D%20%3D0.3%20%5C%5C%20%5C%5C%20%5C%5C%20%20)
Значит, ортофосфат натрия в избытке, а хлорид кальция в недостатке, дальше расчеты ведем по хлориду кальция .
![n(Ca _{3} (PO _{4} ) _{2} )=0.3/3=0.1\\m(Ca _{3} (PO _{4} ) _{2} )=n*M=0.1*(40*3+(31+64)*2)= \\ =0.1*(120+190)=0.1*310=31](https://tex.z-dn.net/?f=n%28Ca%20_%7B3%7D%20%28PO%20_%7B4%7D%20%29%20_%7B2%7D%20%29%3D0.3%2F3%3D0.1%5C%5Cm%28Ca%20_%7B3%7D%20%28PO%20_%7B4%7D%20%29%20_%7B2%7D%20%29%3Dn%2AM%3D0.1%2A%2840%2A3%2B%2831%2B64%29%2A2%29%3D%20%5C%5C%20%3D0.1%2A%28120%2B190%29%3D0.1%2A310%3D31)
[/tex]
<span>S+ Cl2=scl2 2 S+ Cl2=<span>S2Cl2</span></span>
NH3+HCl=NH4CL
n(NH3)= m(NH3) / M(NH3) = 5/17=0.294 моль
n(HCl) =m(HCl) / M(HCl) = 5/36.5 = 0.13698моль
соотношение NH3 и HCl = 1:1
тогда NH3 в избытке
избыток = 0,294-0,13698=0,15702 моль
m(избытка)=n(избытка)*M(NH3)= 0,15702*17=2,669 г