Stoichiometry from oxide reduction: A metal oxide is fully reduced by hydrogen. From 3.15 g of the oxide, 1.05 g of metal is obtained (2.10 g is oxygen). What can be inferred about the metal's equivalent or atomic weight?

Introduction / Context:
Gravimetric data from oxide reduction provide ratios of metal to oxygen that can reveal atomic or equivalent weights. This classic approach predates modern spectroscopy and still trains intuition about formula masses and valency.

Given Data / Assumptions:

• Mass of oxide: 3.15 g → yields metal: 1.05 g; oxygen removed: 2.10 g.
• Atomic mass of oxygen = 16 g/mol.
• Assume the metal forms a simple oxide (commonly MO for divalent metals).

Concept / Approach:
The oxygen-to-metal mass ratio is 2.10 / 1.05 = 2. For an oxide MO, the ratio by mass equals 16 / A_M (A_M = atomic weight of metal). Setting 16 / A_M = 2 gives A_M = 8. While no stable metal has atomic weight 8, the question asks what can be inferred: if the metal were divalent, the equivalent weight E = A_M / valency = 8 / 2 = 4, which matches a valid inference from the data alone (without assigning an element identity).

Step-by-Step Solution:
Compute mass ratio O:M = 2.10 / 1.05 = 2.Assume MO: 16 / A_M = 2 → A_M = 8.For a divalent metal, equivalent weight E = A_M / 2 = 4.Therefore, from the data, we can infer the equivalent weight is 4.

Verification / Alternative check:
If the oxide formula were M2O or M2O3, the inferred atomic weight would differ; however, the simplest consistent interpretation for the provided options is MO leading to E = 4.

Why Other Options Are Wrong:
Atomic weights 4 or 2 conflict with the observed mass ratio; equivalent weight 8 does not follow from MO with valency 2.

Common Pitfalls:

• Confusing atomic with equivalent weight; remember E = A / valency.
• Ignoring valency assumptions embedded in oxide stoichiometry.

Final Answer:
Equivalent weight of the metal is 4