Question

Ammonia (NH3) is one of the most common chemicals produced in the united states. it is used to make fertilizer and other products. Ammonia is produced by the following chemical reaction. N2(g)+3h2(g)= 2nh3(g) If you have 1.00 x 10^3g of N2
a. If you have 1.00 x 10'g of N2 and 2.50 x 10' of H2, which is the limiting reactant in the reaction?
b. How many grams of ammonia can be produced from the amount of limiting reactant available?
c. Calculate the mass of excess reactant that remains after the reaction is complete.

Answer 1

Answer:

a) N2 is the limiting reactant

b) 1215.9 grams NH3

c) 2283.6 grams H2

Explanation:

Step 1: Data given

Mass of N2 = 1000 grams

Molar mass N2 = 28 g/mol

Mass of H2 = 2500 grams

Molar mass H2 = 2.02 g/mol

Step 2: The balanced equation

N2(g) +3H2(g) → 2NH3(g)

Step 3: Calculate moles N2

Moles N2 = mass N2 / molar mass N2

Moles N2 = 1000 grams / 28.0 g/mol

Moles N2 = 35.7 moles

Step 4: Calculate moles H2

Moles H2 = 2500 grams / 2.02 g/mol

Moles H2 = 1237.6 moles

Step 5: Calculate limiting reactant

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

N2 is the limiting reactant. There will be consumed 35.7 moles.

H2 is in excess. There will react 3*35.7 = 107.1 moles

There will remain 1237.6 - 107.1= 1130.5 moles H2

This is 1130.5 * 2.2 = 2283.6 grams H2

Step 6: Calculate moles NH3

For 1 mol N2 we need 3 moles H2 to produce 2 moles NH3

For  35.7 moles N2 we'll have 2*35.7 = 71.4 moles NH3

Step 7: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 71.4 moles * 17.03 g/mol

Mass NH3 = 1215.9 grams NH3