Question

50 kg of N2 gas and 10kg of H2 gas are mixed to produce NH3 gas calculate the NH3gas formed. Identify the limiting reagent in the production of NH3 in this situation

Answer 1

H2 is the limiting reactant

56.2 kg of NH3 will be formed

Step-by-step explanation:

Step 1: Data given

Mass of N2 = 50 kg = 50000 grams

Mass of H2 = 10 kg = 10000 grams

Molar mass of N2 = 28.0 g/mol

Molar mass of H2 = 2.02 g/mol

Molar mass NH3 = 17.03 g/mol

Step 2: The balanced equation

N2(g) + 3H2(g) → 2NH3(g)

Step 3: Calculate moles

Moles = mass / molar mass

Moles N2 = 50000 grams / 28.0 g/mol

Moles N2 = 1785.7 moles

Moles H2 = 10000 grams / 2.02 g/mol

Moles H2 = 4950.5 moles

Step 4: Calculate the limiting reactant

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

H2 is the limiting reactant.It will completely be consumed (4950.5 moles). Né is in excess. There will react 4950.5 / 3 = 1650.2 moles. There will remain 1785.7 - 1650.2 = 135.5 moles

Step 5: Calculate moles NH3

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

For 4950.5 moles H2 we'll have 2/3 * 4950.5 = 3300.3 moles

Step 6: Calculate mass NH3

Mass NH3 = moles NH3 * molar mass NH3

Mass NH3 = 3300.3 moles * 17.03 g/mol

Mass NH3 = 56204 grams = 56.2 kg

Answer 2

1. H2 is the limiting reactant.

2. 56666.67g ( i.e 56.67kg) of NH3 is produced.

Step-by-step explanation:

Step 1:

The equation for the reaction. This is given below:

N2 + H2 —> NH3

Step 2:

Balancing the equation.

N2 + H2 —> NH3

The above equation can be balanced as follow :

There are 2 atoms of N on the left side and 1 atom on the right side. It can be balance by putting 2 in front of NH3 as shown below:

N2 + H2 —> 2NH3

There are 6 atoms of H on the right side and 2 atoms on the left side. It can be balance by putting 3 in front of H2 as shown below

N2 + 3H2 —> 2NH3

Now the equation is balanced.

Step 3:

Determination of the masses of N2 and H2 that reacted and the mass of NH3 produced from the balanced equation. This is illustrated below:

N2 + 3H2 —> 2NH3

Molar Mass of N2 = 2x14 = 28g/mol

Molar Mass of H2 = 2x1 = 2g/mol

Mass of H2 from the balanced equation = 3 x 2 = 6g

Molar Mass of NH3 = 14 + (3x1) = 14 + 3 = 17g/mol

Mass of NH3 from the balanced equation = 2 x 17 = 34g

From the balanced equation above,

28g of N2 reacted with 6g of H2 to produce 34g of NH3

Step 4:

Determination of the limiting reactant. This is illustrated below:

N2 + 3H2 —> 2NH3

Let us consider using all the 10kg (i.e 10000g) of H2 to see if there will be any left of for N2.

From the balanced equation above,

28g of N2 reacted with 6g of H2.

Therefore, Xg of N2 will react with 10000g of H2 i.e

Xg of N2 = (28 x 10000)/6

Xg of N2 = 46666.67g

We can see from the calculations above that there are leftover for N2 as only 46666.67g reacted out of 50kg ( i.e 50000g) that was given. Therefore, H2 is the limiting reactant.

Step 5:

Determination of the mass of NH3 produced during the reaction. This is illustrated below:

N2 + 3H2 —> 2NH3

From the balanced equation above,

6g of H2 reacted to produce 34g of NH3.

Therefore, 10000g of H2 will react to produce = ( 10000 x 34)/6 = 6g of 56666.67g of NH3.

Therefore, 56666.67g ( i.e 56.67kg) of NH3 is produced.