Element Identification and Explanation

To determine the element E based on the information provided, we need to analyze the composition of its higher oxide EO3 and the percentage of hydrogen in its hydride.

Composition of Higher Oxide EO3
The higher oxide of the element E is EO3. This indicates that the element E combines with oxygen to form an oxide with an oxidation state of +3. The formula EO3 suggests that the element E has a valence of +3, meaning it loses three electrons to form a stable compound with oxygen.

Percentage of Hydrogen in Hydride
The given information states that the percentage of hydrogen in the hydride of element E is 2.47%. This suggests that the hydride formula is EHx, where x represents the number of hydrogen atoms.

Analysis and Deduction
To determine the valence of element E, we need to compare the number of electrons gained or lost with the number of hydrogen atoms in the hydride.

1. The formula of the higher oxide EO3 indicates that element E loses three electrons to attain a stable oxidation state of +3. This suggests that E has three valence electrons.

2. The percentage of hydrogen in the hydride is 2.47%. This implies that the number of hydrogen atoms is 2.47% of the total mass of the hydride.

3. To find the molar mass of the hydride, we can assume a convenient mass for the hydride, such as 100 grams. Therefore, the mass of hydrogen in the hydride would be 2.47 grams (2.47% of 100 grams).

4. The molar mass of hydrogen is approximately 1 gram/mol. So, the number of moles of hydrogen in the hydride is 2.47 moles (2.47 grams / 1 gram/mol).

5. Since the hydride formula is EHx, the number of moles of hydrogen should be equal to x. Therefore, we deduce that x = 2.47.

Conclusion
Based on the analysis, the element E has a valence of +3 and the hydride formula EH2. Therefore, the element E is likely to be boron (B) since it forms the higher oxide B2O3 and the hydride BH3.