Respuesta :
Answer:
1. H2 is the limiting reactant.
2. 56666.67g ( i.e 56.67kg) of NH3 is produced.
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.
Answer:
H2 is the limiting reactant
56.2 kg of NH3 will be formed
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
