H2 is limiting reagent and O2 is excess reagent and 36 gm of water is formed

Analysis:
To solve this problem, we need to determine the limiting reagent, calculate the maximum amount of water formed, and find the amount of excess reactant left unreacted. This can be done by comparing the stoichiometry of the reaction and using the concept of limiting reagents.

Solution:

1. Determining the Limiting Reagent:
To determine the limiting reagent, we need to compare the amount of each reactant (H2 and O2) with their stoichiometric coefficients in the balanced equation.

The balanced equation for the reaction is:
2H2 + O2 → 2H2O

Given:
Mass of H2 = 3g
Mass of O2 = 29g

Using the molar mass of H2 (2g/mol) and O2 (32g/mol), we can calculate the number of moles for each reactant:

Number of moles of H2 = Mass of H2 / Molar mass of H2 = 3g / 2g/mol = 1.5 mol
Number of moles of O2 = Mass of O2 / Molar mass of O2 = 29g / 32g/mol ≈ 0.90625 mol

Now, we need to compare the moles of reactants to their stoichiometric coefficients in the balanced equation. The stoichiometric coefficients tell us the mole ratio between reactants and products.

According to the balanced equation, the mole ratio of H2 to O2 is 2:1. This means that for every 2 moles of H2, 1 mole of O2 is required.

2. Calculating the Maximum Amount of Water Formed:
Since the balanced equation shows that 2 moles of H2 react with 1 mole of O2 to produce 2 moles of H2O, we can use stoichiometry to calculate the maximum amount of water formed.

From the given information, we know that the number of moles of O2 is 0.90625 mol. According to the stoichiometry, this amount of O2 would require 2 * 0.90625 = 1.8125 moles of H2.

Since we have only 1.5 moles of H2, which is less than the required amount, H2 is the limiting reagent. The maximum amount of water formed is determined by the limiting reagent.

To calculate the maximum amount of water formed, we use the stoichiometry of the balanced equation. According to the equation, 2 moles of H2 produce 2 moles of H2O. Therefore, 1.5 moles of H2 will produce (1.5 * 2) = 3 moles of H2O.

To convert moles of H2O to grams, we use the molar mass of water (18g/mol):

Mass of water = Number of moles of water * Molar mass of water
Mass of water = 3 moles * 18g/mol = 54g

Hence, the maximum amount of water formed is 54g.

3. Calculating the Excess Reactant:
The excess reactant is the reactant that is not completely consumed in the reaction. In this case, the excess reactant is O2