Ammonia gas reacts with oxygen gas according to the following equation:

4NH3 + 5O2----4NO + 6H2O
a. How many moles of oxygen gas are needed to react with 23 moles of ammonia? (29 mole)
b. How may grams of NO are produced when 25 moles of oxygen gas react with an excess of ammonia? (600g)
c. If 24 grams of water are produced, how many moles of nitrogen monoxide are formed? (0.89 mole)
d. How many grams of oxygen are need to react with 6.78 grams of ammonia? (16.0 g)

Pio

Calculate the number of moles of nitrogen monoxide needed for 2.5 moles of oxygen to react

2d

To solve these problems, we need to use the stoichiometry of the chemical reaction, which relates the amounts of reactants and products in a balanced equation.

a. To determine how many moles of oxygen gas are needed to react with 23 moles of ammonia, we can use the coefficients in the balanced equation. According to the equation, the ratio of ammonia to oxygen is 4:5.
So, we can set up a proportion: (4 moles of NH3 / 5 moles of O2) = (23 moles of NH3 / x moles of O2).

To find x, we cross multiply and solve for x:
4x = 5 * 23
4x = 115
x = 115 / 4
x = 28.75

Therefore, we need approximately 29 moles of oxygen gas to react with 23 moles of ammonia.

b. To determine the grams of NO produced when 25 moles of oxygen gas react with an excess of ammonia, we need to use the molar mass of NO. According to the balanced equation, the molar ratio of oxygen gas to NO is 5:4.

First, calculate the number of moles of NO produced:
(25 moles of O2) * (4 moles of NO / 5 moles of O2) = 20 moles of NO

Then, we can use the molar mass of NO to convert to grams:
(20 moles of NO) * (30.01 g/mol) = 600.2 grams of NO (approx. 600 g)

Therefore, approximately 600 grams of NO are produced when 25 moles of oxygen gas react with an excess of ammonia.

c. To determine the number of moles of nitrogen monoxide formed when 24 grams of water are produced, we need to reverse the calculation from part b. We will use the molar mass of water (H2O) and the molar ratio of water to NO (6:4) from the balanced equation.

First, calculate the number of moles of NO produced:
(24 g of H2O) * (1 mol/18.02 g) * (4 mol NO/6 mol H2O) = 0.89 moles of NO

Therefore, 0.89 moles of nitrogen monoxide are formed when 24 grams of water are produced.

d. To determine the grams of oxygen needed to react with 6.78 grams of ammonia, we need to use the molar mass of ammonia (NH3) and the molar ratio of ammonia to oxygen from the balanced equation (4:5).

First, calculate the number of moles of ammonia:
(6.78 g of NH3) * (1 mol/17.03 g) = 0.398 moles of NH3

Then, we can use the mole ratio to find the moles of oxygen needed:
(0.398 moles of NH3) * (5 moles of O2/4 moles of NH3) = 0.4975 moles of O2

Finally, convert moles of O2 to grams using the molar mass of oxygen (32.00 g/mol):
(0.4975 moles of O2) * (32.00 g/mol) = 15.92 grams (approx. 16.0 g)

Therefore, approximately 16.0 grams of oxygen are needed to react with 6.78 grams of ammonia.

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