In an experiment, 3.25 g of NH3 are allowed to react with 3.50 g of O2. 4NH3 + 5O2 > 4NO + 6H2O a. Which reactant is the limiting reagent? O2 b. How many grams of NO are formed

2.61 g of NO will be formed

The limiting reagent is the O₂

Explanation:

The reaction is:

4NH₃  +  5O₂  →  4NO  +  6H₂O

We convert the mass of the reactants to moles:

3.25g / 17 g/mol = 0.191 moles of NH₃

3.50g / 32 g/mol =0.109 moles of O₂

Let's determine the limiting reactant by stoichiometry:

4 moles of ammonia react with 5 moles of oxygen

Then, 0.191 moles of ammonia will react with (0.191 . 5) / 4 = 0.238 moles of oxygen. We only have 0.109 moles of O₂ and we need 0.238, so as the oxygen is not enough, this is the limiting reagent

Ratio with NO is 5:4

5 moles of oxygen produce 4 moles of NO

0.109 moles will produce (0.109 . 4)/ 5 = 0.0872 moles of NO

We convert the moles to mass, to get the answer

0.0872 mol . 30g / 1 mol = 2.61 g

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be

catya that's not even a b or c