25 g of NH, is mixed with 4 moles of O, in the given reaction: 4NH,(g) + 50 (8) 4NO(g) + 6H,O(1)
a Which is the limiting reactant?
b What mass of NO is formed?
c What mass of H2O is formed?

a. NH3 is limiting reactant.

b. 44g of NO

c. 40g of H2O

Explanation:

Based on the reaction:

4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(l)

4 moles of ammonia reacts with 5 moles of oxygen to produces 4 moles of NO and 6 moles of water.

To find limiting reactant we need to find the moles of each reactant and using the balanced equation find which reactant will be ended first. Then, with limiting reactant we can find the moles of each reactant and its mass:

a. Moles NH3 -Molar mass. 17.031g/mol-

25g NH3*(1mol/17.031g) = 1.47moles NH3

Moles O2 = 4 moles

For a complete reaction of 4 moles of O2 are required:

4mol O2 * (4mol NH3 / 5mol O2) = 3.2 moles of NH3.

As there are just 1.47 moles, NH3 is limiting reactant

b. Moles NO:

1.47moles NH3 * (4mol NO/4mol NH3) = 1.47mol NO

Mass NO -Molar mass: 30.01g/mol-

1.47mol NO * (30.01g/mol) = 44g of NO

c. Moles H2O:

1.47moles NH3 * (6mol H2O/4mol NH3) = 2.205mol H2O

Mass H2O -Molar mass: 18.01g/mol-

2.205mol H2O * (18.01g/mol) = 40g of H2O

a) NH₃ is the limiting reactant

b) 44.1 g NO

c) 39.7 g H₂O

Explanation:

The chemical equation is the following:

4 NH₃(g) + 5 O₂(g) → 4 NO(g) + 6 H₂O(l)

According to the balanced chemical equation, 4 moles of NH₃ reacts with 5 moles of O₂ to produce 4 moles of NO and 6 moles of H₂O.

a) To determine which is the limiting reactant we have to compare the stoichiometric amounts of reactants (NH₃ and O₂) with the actual amounts.

Stoichiometric amounts:

Molecular weight of NH₃= 14 g/mol + (1 g/mol x 3) = 17 g/mol

Stiochiometric amount of NH₃ = 4 moles NH₃ x 17 g/mol = 68 g NH₃

Stiochiometric amount of O₂ = 5 moles

⇒ Stiochiometric ratio = 68 g NH₃/5 moles O₂

Then, we multiply the stoichiometric ratio by the actual amount of O₂ to calculate the required amount of NH₃:

required NH₃ = 4 moles O₂ x 68 g NH₃/5 moles O₂ = 54.4 g NH₃

Thus, we need 54.4 grams of NH₃ to completely react with 4 moles of O₂, but we only have 25 grams of NH₃. Therefore, NH₃ is the limiting reactant.

b) To calculate the amount of NO formed, we use the limiting reactant (NH₃). According to the equation, 4 moles of NH₃ produce 4 moles of NO.

Molecular weight NO = 14 g/mol + 16 g/mol = 30 g/mol

4 mol NH₃ x 17 g/mol = 68 g NH₃

4 mol NO x 30 g/mol = 120 g NO

⇒ Stiochiometric ratio = 120 g NO/68 g NH₃

Now, we multiply the stoichiometric ratio by the actual amount of NH₃ to calculate the mass of NO formed:

NO formed = 25 g NH₃ x 120 g NO/68 g NH₃ = 44.1 g NO

c) According to the chemical equation, 4 moles of NH₃ (68 g) produce 6 moles of H₂O.

Molecular weight H₂O = (2 x 1 g/mol) + 16 g/mol = 18 g/mol

6 mol H₂O x 18 g/mol = 108 g H₂O

⇒ Stoichiometric ratio = 108 g H₂O/68 g NH₃

Finally, we multiply the stoichiometric ratio by the actual amount of NH₃ (limiting reactant) to obtain the mass of H₂O formed:

H₂O formed = 25 g NH₃ x 108 g H₂O/68 g NH₃ = 39.7 g H₂O

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Explanation:

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