Odpowiedzi

2010-02-17T10:03:55+01:00
Równanie reakcji;
3Mg+2H3PO4=Mg3(PO4)2+3H2
a)
3*24g(Mg)- 3mole(H2)
7,2g(Mg)-x(moli H2)
x=3mole*7,2g/3*24g
x=0,3mola

b)
3*24g(Mg)- 6g(H2)
7,2g(Mg)-x(gH2)
x=6g*7,2g/3*24g
x=0,6g

c)
3*24g(Mg)- 3*6,02*10-23(H2)
7,2g(Mg)-x(cząsteczek H2)
x=3*6,02*10-23cząst*7,2g/3*24g
x=1,806*10-23cząsteczek wodoru

d)
3*24g(Mg)- 3*22,4dm3(H2)
7,2g(Mg)-x(dm3 H2)
x=3*22.4dm3*7,2g/3*24g
x=6,72dm3
2010-02-17T10:20:11+01:00
Witaj
3Mg.......+.......2H3PO4......=........Mg2(PO4)2.......+..........3H2
72g Mg.................................................................3 mole H2
7,2g Mg...................................................................x
-------------------------------------------------------------------------------------
x = 7,2g*3mole/72g = 0,3 mola H2

72gMg.............6g H2
7,2...................y
----------------------------
y = 0,6g H2

72g Mg.............67,2dm3 H2
7,2.....................V
-------------------------------------
V = 6,72dm3 H2

72g Mg..............3N cząsteczek H2 N - liczba Avogadro = 6,02*10^23
7,2g....................n
-------------------------------
n = 0,3N = 0,3*6,02*10^23 = 1,806*10^23 cząsteczek H2

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