A 2.00 g sample of ammonia is mixed with 4.00 g of oxygen. Which is thelimiting reactant and how much excess reactant remains after the reaction
has stopped?

Answer:

Ammonia is limiting reactant

Amount of oxygen left  = 0.035 mol

Explanation:

Given data:

Masa of ammonia = 2.00 g

Mass of oxygen = 4.00 g

Which is limiting reactant = ?

Excess reactant's amount left = ?

Solution:

Balance chemical equation:

4NH₃ + 3O₂     →     2N₂ + 6H₂O

Number of moles of ammonia:

Number of moles = mass/molar mass

Number of moles = 2.00 g/ 17 g/mol

Number of moles = 0.12 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 4.00 g/ 32 g/mol

Number of moles = 0.125 mol

Now we will compare the moles of ammonia and oxygen with water and nitrogen.

                       NH₃          :            N₂

                         4             :             2

                       0.12           :           2/4×0.12 = 0.06

                       NH₃         :            H₂O

                         4            :             6

                         0.12       :           6/4×0.12 = 0.18

                        O₂            :            N₂

                         3             :             2

                       0.125        :           2/3×0.125 = 0.08

                         O₂           :            H₂O

                         3              :             6

                         0.125       :           6/3×0.125 = 0.25

The number of moles of water and nitrogen formed by ammonia are less thus ammonia will be limiting reactant.

Amount of oxygen left:

                         NH₃          :             O₂

                            4            :              3

                            0.12       :          3/4×0.12= 0.09

Amount of oxygen react = 0.09 mol

Amount of oxygen left  = 0.125 - 0.09 = 0.035 mol

Final answer:

The limiting reactant in the reaction between Ammonia and Oxygen is Ammonia (NH3). All of the Oxygen is used up in the reaction, so no excess reactant remains.

Explanation:

This question involves a concept in chemistry known as limiting reactants and stoichiometry. The balanced chemical reaction between Ammonia (NH3) and Oxygen (O2) is: 4NH3 + 5O2 -> 4NO + 6H2O. This indicates that 4 moles of NH3 react with 5 moles of O2.

To find the limiting reactant, you first need to convert the grams of your reactants to moles. The molar mass of NH3 is approximately 17.0g/mol, and the molar mass of O2 is 32.0g/mol. Therefore, you have 2.00g/17.0g/mol = 0.118 moles of NH3 and 4.00g/32.0g/mol = 0.125 moles of O2.

Since 5 moles of O2 are needed for every 4 moles of NH3, and we have slightly more O2 than NH3, the limiting reactant is NH3. To find the amount of excess reactant, we determine how much O2 actually reacted by multiplying (0.118 moles NH3)*(5 moles O2/4 moles NH3) = 0.1475 moles O2. The original amount of O2 was 0.125 moles, so the amount left over is 0.125 - 0.1475, which is a negative number and thus not possible. This confirms that O2 is the excess reactant, although it entirely reacted. Hence, no excess reactant remains.

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