Answer:
Explanation:
Mass of O / Mass of Fe = .2865/1 = 112 x .2865/112 = 32.088/112
Mass of O / Mass of Fe = .4297 /1 = 112 x .4297 / 112 = 48 / 112
Ratio of mass of O That reacts with constant mass of Fe of 112 g is as follows
32.088 : 48 = 32/16 : 48/16
= 2 : 3
Hence given data are consistent with the law of multiple proportions.
Answer:
- For FeO:
[tex]\frac{O}{Fe}=\frac{16}{55.8}=0.2865[/tex]
- For Fe₂O₃:
[tex]\frac{O}{Fe}=\frac{16*3}{55.8*2}=0.43[/tex]
Explanation:
Hello,
The law of multiple proportions states that: "if two elements form more than one compound between them, then the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers". In this manner, both FeO and Fe₂O₃ are in such way that they have the small whole numbers in their structures, thus, to compute the required ratios we perform the following mathematical relationships:
- For FeO:
[tex]\frac{O}{Fe}=\frac{16}{55.8}=0.2865[/tex]
- For Fe₂O₃:
[tex]\frac{O}{Fe}=\frac{16*3}{55.8*2}=0.43[/tex]
Wherein such values are consistent with the given data.
Best regards.
