Question

2. 14g of Nitrogen gas and 8.0g of hydrogen react to produce ammonia according to the equation: N2 + 3H2 -- 2NH3 Calculate the mass of hydrogen leftover once the reaction has happened.
a) Identify the limiting reagent in this reaction. Give a reason for your answer.
b) Determine the amount, in moles of the limiting reagent.
c) Determine the amount, in moles of the excess reagent.
d) Determine which reactant will produce the least amount of ammonia.
e) Calculate the amount, in moles of H2, reacted, when the limiting reagent has been used up.
f) Give the mass of the amount of H2 that has reacted​

Answer 1

Based on the equation of the reaction, nitrogen is the limiting reagent while hydrogen is the excess reagent.

What is the mole ratio of hydrogen to nitrogen in the formation of ammonia?

Hydrogen and nitrogen combines to form ammonia ina mole ratio of 3 : 1 as shown by the equation of the reaction below:

• N2 + 3H2 -- 2NH3

The number of moles of the reactants in 14g of Nitrogen gas and 8.0g of hydrogen is calculated as follows:

• Moles = mass/molar mass

Molar mass of N_{2} = 14.0 g

Molar mass of H_{2} = 2.0 g

Moles of N_{2} = 14/14.0 = 1 mole

Moles of H_{2} = 8/2.0 = 4 moles

Based on the data above:

• The limiting reagent is nitrogen gas as it will be used up while hydrogen will be left over.
• The moles of nitrogen is 1 mole
• Hydrogen is the excess reagent and 1 mole will be left over
• 3 moles of hydrogen will react with 1 mole of the nitrogen gas
• mass of 3 moles of hydrogen is 3 × 2.0 g = 6.0g

Therefore, the limiting reagent is nitrogen while hydrogen is the excess reagent.

Learn more about limiting reagent at: brainly.com/question/24945784