Q1: A metal oxide is reduced by heating it in a stream of hydrogen. After complete reduction, it is found that 3.15 gm of the oxide has yielded 1.05 gm of the metal. It may be inferred that the

ANS:B - equivalent weight of the metal is 4.

To solve this problem, we need to understand the concept of stoichiometry, which involves the quantitative relationships between reactants and products in a chemical reaction. When a metal oxide (M𝑥O𝑦Mx​Oy​) is reduced by hydrogen gas (H2H2​), the following reaction occurs: M𝑥O𝑦+𝑦H2→𝑥M+𝑦H2OMx​Oy​+yH2​→xM+yH2​O Given that 3.15 grams of the metal oxide yields 1.05 grams of the metal, we can set up a proportion to find the ratio of the masses of the metal to the metal oxide: Mass of metalMass of metal oxide=1.053.15Mass of metal oxideMass of metal​=3.151.05​ Now, we need to determine the atomic weight or equivalent weight of the metal. Since we're dealing with an oxide, we'll need to consider the stoichiometry of the reaction. From the balanced equation, we can see that 1 mol1 mol of the metal oxide reacts with 𝑦y mols of hydrogen gas to produce 𝑥x mols of metal. So, 1 mol1 mol of the metal oxide yields 𝑥 molsx mols of the metal. Using the given masses, we can convert them into moles: moles of metal oxide=mass of metal oxidemolar mass of metal oxidemoles of metal oxide=molar mass of metal oxidemass of metal oxide​ moles of metal=mass of metalmolar mass of metalmoles of metal=molar mass of metalmass of metal​ Since the molar mass of the metal oxide (M𝑥O𝑦Mx​Oy​) is unknown, let's denote it as 𝑀oxideMoxide​, and the molar mass of the metal as 𝑀metalMmetal​. So, according to the stoichiometry of the reaction: moles of metalmoles of metal oxide=𝑥1moles of metal oxidemoles of metal​=1x​ mass of metal𝑀metalmass of metal oxide𝑀oxide=𝑥1Moxide​mass of metal oxide​Mmetal​mass of metal​​=1x​ Given that 3.15 grams of the oxide yields 1.05 grams of the metal: 1.05𝑀metal3.15𝑀oxide=𝑥Moxide​3.15​Mmetal​1.05​​=x Now, let's solve for 𝑥x by substituting the given values: 1.05𝑀metal3.15𝑀oxide=𝑥Moxide​3.15​Mmetal​1.05​​=x 𝑥=1.053.15×𝑀oxide𝑀metalx=3.151.05​×Mmetal​Moxide​​ Given that 𝑥=1x=1, because one mole of metal oxide yields one mole of metal, we can solve for 𝑀oxide𝑀metalMmetal​Moxide​​: 1=1.053.15×𝑀oxide𝑀metal1=3.151.05​×Mmetal​Moxide​​ 𝑀oxide=3.15×𝑀metal1.05Moxide​=3.15×1.05Mmetal​​ Given that the molar mass of the metal oxide (M𝑥O𝑦Mx​Oy​) is 𝑀oxideMoxide​ grams, and the molar mass of the metal is 𝑀metalMmetal​ grams, the ratio 𝑀oxide𝑀metalMmetal​Moxide​​ represents the stoichiometry of the reaction, which is often denoted by the term "equivalent weight." Therefore, the equivalent weight of the metal is 3.15 times the molar mass of the metal divided by 1.05. Let's calculate it. Given that the molar mass of the metal oxide (M𝑥O𝑦Mx​Oy​) is 𝑀oxideMoxide​ grams and that of the metal is 𝑀metalMmetal​ grams, we can write: 𝑀oxide=3.15×𝑀metal1.05Moxide​=3.15×1.05Mmetal​​ Given the options, we are asked to find if the atomic weight of the metal is 4 or 2, or if the equivalent weight of the metal is 4 or 8. Let's start by calculating the equivalent weight of the metal: Equivalent weight of the metal=𝑀oxide1Equivalent weight of the metal=1Moxide​​ Equivalent weight of the metal=3.15×𝑀metal1.05Equivalent weight of the metal=3.15×1.05Mmetal​​ Given that 3.15 grams of the oxide yields 1.05 grams of the metal: Equivalent weight of the metal=3.15×𝑀metal1.05Equivalent weight of the metal=3.15×1.05Mmetal​​ Equivalent weight of the metal=3.15×𝑀metal1.05Equivalent weight of the metal=3.15×1.05Mmetal​​ Equivalent weight of the metal=3×𝑀metal1.05Equivalent weight of the metal=3×1.05Mmetal​​ Equivalent weight of the metal=31.05×𝑀metalEquivalent weight of the metal=1.053​×Mmetal​ Equivalent weight of the metal=6021×𝑀metalEquivalent weight of the metal=2160​×Mmetal​ Equivalent weight of the metal=207×𝑀metalEquivalent weight of the metal=720​×Mmetal​ Equivalent weight of the metal≈2.857×𝑀metalEquivalent weight of the metal≈2.857×Mmetal​ So, the equivalent weight of the metal is approximately 2.857×𝑀metal2.857×Mmetal​. Given that 𝑀metalMmetal​ is a constant, the equivalent weight of the metal will vary accordingly. None of the options exactly match the calculated equivalent weight, but it is closest to 8. Therefore, we can infer that the equivalent weight of the metal is approximately 8.