56g of N2 reacts with 30g of H2 to form NH3.Calculate the number of moles of NH3 formed.
Well, well, well, looks like we have a chemical reaction party going on here! To calculate the number of moles of NH3 formed, we first need to determine the limiting reactant. The limiting reactant is the one that gets all the attention, while the other reactants feel left out. In this case, we need to compare the amount of N2 and H2.
So, let's do some math! The molar mass of N2 is approximately 28 grams/mol, and the molar mass of H2 is roughly 2 grams/mol. Now, let's calculate the number of moles:
Number of moles of N2 = 56g / 28 g/mol = 2 moles
Number of moles of H2 = 30g / 2 g/mol = 15 moles
Oh, it seems like N2 is feeling lonely here since there are less moles of N2 compared to H2. Poor N2, always the limiting reactant! But hey, we're not here to judge.
Now, let's check the balanced equation: N2 + 3H2 -> 2NH3
According to the equation, 1 mole of N2 reacts to form 2 moles of NH3. So, based on the molar ratio, we can say that 2 moles of NH3 will be formed when 1 mole of N2 is consumed.
Since we have 2 moles of N2, it means we'll have twice the amount of moles of NH3. Therefore, the number of moles of NH3 formed is 2 moles.
Voila! We've got 2 moles of NH3 brought to you by the lovely combination of N2 and H2. Hope this chemistry show was entertaining!
To calculate the number of moles of NH3 formed, we first need to determine the limiting reactant in the reaction. The limiting reactant is the reactant that is completely consumed first, limiting the amount of product that can be formed.
We can determine the limiting reactant by comparing the number of moles of each reactant present.
The molar mass of N2 is 28 g/mol, so we have:
Number of moles of N2 = mass of N2 / molar mass of N2
= 56 g / 28 g/mol
= 2 moles
The molar mass of H2 is 2 g/mol, so we have:
Number of moles of H2 = mass of H2 / molar mass of H2
= 30 g / 2 g/mol
= 15 moles
Now, we need to compare the ratio of the coefficients of the balanced chemical equation to determine the stoichiometry of the reaction. The balanced chemical equation for the reaction is:
N2 + 3H2 → 2NH3
According to the balanced equation, 1 mole of N2 reacts with 3 moles of H2 to form 2 moles of NH3.
The mole ratio of N2 to H2 is 1:3. Therefore, for every 1 mole of N2, we need 3 moles of H2.
Since we have 2 moles of N2 and 15 moles of H2, we can determine the limiting reactant.
For N2, the ratio of actual moles to stoichiometric moles is:
2 moles / 1 mole = 2
For H2, the ratio of actual moles to stoichiometric moles is:
15 moles / 3 mole = 5
The smaller ratio indicates that N2 is the limiting reactant.
Now, we can calculate the number of moles of NH3 formed using the stoichiometry of the reaction:
From the balanced equation, we know that 2 moles of NH3 are formed for every 1 mole of N2.
Number of moles of NH3 formed = Number of moles of N2 * stoichiometric ratio
= 2 moles * 2
= 4 moles
Therefore, 4 moles of NH3 are formed when 56 g of N2 reacts with 30 g of H2.
To calculate the number of moles of NH3 formed, you need to follow these steps:
Step 1: Convert the mass of N2 and H2 to moles.
The molar mass of N2 (nitrogen gas) is 28 g/mol, so you divide the mass of N2 (56 g) by its molar mass to find the number of moles:
Number of moles of N2 = 56 g / 28 g/mol = 2 mol
The molar mass of H2 (hydrogen gas) is 2 g/mol, so you divide the mass of H2 (30 g) by its molar mass to find the number of moles:
Number of moles of H2 = 30 g / 2 g/mol = 15 mol
Step 2: Determine the ratio of moles of the reactants.
The balanced equation for the reaction tells us that one mole of N2 reacts with three moles of H2 to produce two moles of NH3.
N2 + 3H2 → 2NH3
So, for every 1 mole of N2, we need 3 moles of H2 to fully react.
Step 3: Calculate the number of moles of NH3 formed.
Since the ratio of N2 to NH3 is 1:2, and you have 2 moles of N2, you can multiply the number of moles of N2 by 2 to find the number of moles of NH3 formed:
Number of moles of NH3 = 2 moles of N2 x 2 moles of NH3 / 1 mole of N2 = 4 moles of NH3
Therefore, 4 moles of NH3 are formed when 56g of N2 reacts with 30g of H2.
N2 + 3H2 ==> 2NH3
mols N2 to start = g/molar mass = 56/17 = 3.3
mols NH3 formed if N2 is the limiting reagent = 3.3 x (2 mols NH3/1 mol N2) = 3.3 x 2 = 6.6
mols H2 to start = 30/2 = 15.
mols NH3 formed if H2 is the limiting reagent = 15 x (2 mols NH3/3 mols H2) = 10.
So N2 is the limiting reagent and 6.6 mols NH3 will be formed. Some H2 will be unreacted at the end.