El perclorato de potacio KClO4 se descompone produciendo KCl y O2, Segun la ecuacion KClO4 ---> KCl + 2O2 a) Calcule la masa del KClO4 requerida para producir 1.5 gramos de O2b) Calcule la masa del KCl que se produce coln este oxigenoc) Calcule cuantos gramos de KClO4 se requiere para producir 97 gramos de O2
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KClO4 → KCl + 2 O2
Mm KClO4 = 138.54 g/mol
KCl = 74.55 g/mol
O2 = 32 g/mol
a) g de KClO4 = ?
g = 1.5 g O2 x 1 mol O2 x 1 mol KClO4 x 138.54 g KClO4
`````````````` ``````````````````` `````````````````````````
32 g O2 2 mol O2 1 mol KClO4
g = 3.25 g KClO4
b) gramos de KCl
g = 1.5 g O2 x 1 mol O2 x 1 mol KCl x 74.55 g KCl
`````````````` ``````````````````` ````````````````````
32 g O2 2 mol O2 1 mol KCl
g = 1.75 g KCl
c) g KClO4 = ?
g = 97 g O2 x 1 mol O2 x 1 mol KClO4 x 138.54 g KClO4
`````````````` ``````````````````` `````````````````````````
32 g O2 2 mol O2 1 mol KClO4
g = 209.97 g KClO4
Mm KClO4 = 138.54 g/mol
KCl = 74.55 g/mol
O2 = 32 g/mol
a) g de KClO4 = ?
g = 1.5 g O2 x 1 mol O2 x 1 mol KClO4 x 138.54 g KClO4
`````````````` ``````````````````` `````````````````````````
32 g O2 2 mol O2 1 mol KClO4
g = 3.25 g KClO4
b) gramos de KCl
g = 1.5 g O2 x 1 mol O2 x 1 mol KCl x 74.55 g KCl
`````````````` ``````````````````` ````````````````````
32 g O2 2 mol O2 1 mol KCl
g = 1.75 g KCl
c) g KClO4 = ?
g = 97 g O2 x 1 mol O2 x 1 mol KClO4 x 138.54 g KClO4
`````````````` ``````````````````` `````````````````````````
32 g O2 2 mol O2 1 mol KClO4
g = 209.97 g KClO4
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