Răspuns:
Explicație:
H3PO4
a.
acid fosforic
b.
r.a.= 3 : 1 : 4
r.m. = 3 : 31 : 4.16
= 3 : 31 : 64
c.
comp. procentuala
M H3PO4= 3 + 31 + 64=98 -------> 98g/moli
98g H3PO4 -----------------64 g O----3gH----------31gP
100g------------------------------x-------------y----------------z
x= 100 . 64 : 98 = 63 , 31 % O
y=3,06% H
z = 31,63% P
d.
5moli H3PO4
nr. atomi H
1mol H3PO4---------------3 . 6,023.10²³ atomi H
5moli H3PO4-----------------------x
x= 5 . 3 . 6,023.10²³atomi H = 90,345 .10²³ atomi H
e.
n=nr. moli
n = 120,44.10 ²³ : 6,023.10²³= 19,99 moli H3PO4
m= n . M
m= 19,99 moli . 98g/moli = 1959,02 g H3PO4
aproximativ ---> 1960g
f.
MH3PO4= 98----> 98kg/kmoli
98 kg H3PO4--------64 kg O
xkg------------------------6,4kg
x= 98 . 6,4 : 64 = 9,8 kg H3PO4 = 9800g
n= 9,8 kg : 98kg/kmoli= 0,1 kmoli = 100moli H3PO4
