1.
Mg_{3}P_{2} ->PH_{3} ->P_{2}O_{5}-> H_{3}PO_{4} -> Ca_{3}(PO_{4})_{2} " alt=" P ->Mg_{3}P_{2} ->PH_{3} ->P_{2}O_{5}-> H_{3}PO_{4} -> Ca_{3}(PO_{4})_{2} " align="absmiddle" class="latex-formula">
1) ![2P +3Mg=Mg_{3}P_{2} 2P +3Mg=Mg_{3}P_{2}](https://tex.z-dn.net/?f=2P+%2B3Mg%3DMg_%7B3%7DP_%7B2%7D+)
2) ![Mg_{3}P_{2}+6HCl=3MgCl_{2}+2PH_{3} Mg_{3}P_{2}+6HCl=3MgCl_{2}+2PH_{3}](https://tex.z-dn.net/?f=Mg_%7B3%7DP_%7B2%7D%2B6HCl%3D3MgCl_%7B2%7D%2B2PH_%7B3%7D+)
3) ![2PH_{3}+4O_{2}=P_{2}O_{5}+3H_{2}O 2PH_{3}+4O_{2}=P_{2}O_{5}+3H_{2}O](https://tex.z-dn.net/?f=2PH_%7B3%7D%2B4O_%7B2%7D%3DP_%7B2%7DO_%7B5%7D%2B3H_%7B2%7DO+)
4) ![P_{2}O_{5}+3H_{2}O=2H_{3}PO_{4} P_{2}O_{5}+3H_{2}O=2H_{3}PO_{4}](https://tex.z-dn.net/?f=P_%7B2%7DO_%7B5%7D%2B3H_%7B2%7DO%3D2H_%7B3%7DPO_%7B4%7D+)
5) ![2H_{3}PO_{4}+3Ca=Ca_{3}(PO_{4})_{2}+3H_{2} 2H_{3}PO_{4}+3Ca=Ca_{3}(PO_{4})_{2}+3H_{2}](https://tex.z-dn.net/?f=2H_%7B3%7DPO_%7B4%7D%2B3Ca%3DCa_%7B3%7D%28PO_%7B4%7D%29_%7B2%7D%2B3H_%7B2%7D+)
2.
3H_{3}PO_{4}+5NO " alt="3P+5HNO_{3}+2H_{2}O->3H_{3}PO_{4}+5NO " align="absmiddle" class="latex-formula">
окисление
P^{+5} " alt="P^{0}->P^{+5} " align="absmiddle" class="latex-formula"> (-5e) восстановитель |5|3
восстановление
N^{+2}" alt="N^{+5}->N^{+2}" align="absmiddle" class="latex-formula"> (+3e) окислитель |3|5
3. Дано:
=80%
m(P)=31кг
=5%
________________________________
m
=?
Решение:
0,2 x
3H_{3}PO_{4}+5NO " alt="3P+5HNO_{3}+2H_{2}O->3H_{3}PO_{4}+5NO " align="absmiddle" class="latex-formula">
3кмоль 3кмоль
чист.(P)=100%-80%=20%
mчист.(P)=31кг*0,2=6,2кг
n(P)=6,2кг/31кг/кмоль=0,2кмоль
x=n(
)=0,2кмоль
m(
)=0,2кмоль*98кг/кмоль=19,6кг
m(
)=19,6кг*08=15,68 кг
Ответ:15,68кг