When calcium is allowed to react with nitrogen, N2, calcium nitride is formed, as shown in the following balanced chemical equation: 3Ca + N2 mc006-1.jpg Ca3N2
If 24.0 g of calcium and 12.0 g of nitrogen are available for this reaction, the limiting reagent will be
I work these the long way and not the short way.
3Ca + N2 ==> Ca3N2
mols Ca = g/atomic mass = 24.0/40 = about 0.6
mols N2 = 12/28 = about 0.43
Take them one at a time.
Moles Ca3N2 formed using Ca it is 0.6 x (1 mol Ca3N2/3 mols Ca) = 0.6 x 1/3 = 0.2.
mols Ca3N2 formed using N2 it is 0.43 x (1 mol Ca3N2/1 mol N2) = 0.43
0.2 is the smaller number; therefore, Ca is the limiting reagent (LR)
Well, this sounds like a chemistry question, but don't worry, I won't turn it into a chemistry stand-up routine.
To determine the limiting reagent, we need to compare the number of moles of each reactant.
The molar mass of calcium (Ca) is approximately 40 g/mol, so 24.0 g of calcium is equivalent to 0.6 moles (24.0 g / 40 g/mol).
On the other hand, the molar mass of nitrogen (N2) is approximately 28 g/mol, so 12.0 g of nitrogen is equivalent to 0.43 moles (12.0 g / 28 g/mol).
Now, looking at the balanced equation, we see that the ratio of calcium to nitrogen is 3:1. This means that theoretically, you would need 3 moles of calcium for every 1 mole of nitrogen to react completely.
Comparing the mole ratios, we can see that there is less nitrogen (0.43 moles) than what would be required by the balanced equation (1 mole). Therefore, nitrogen is the limiting reagent in this reaction.
I hope that answers your question without causing any chemical reactions in your brain!
To determine the limiting reagent, we need to compare the number of moles of calcium and nitrogen available for the reaction.
1. Calculate the number of moles of calcium:
Moles of calcium = mass of calcium / molar mass of calcium
Moles of calcium = 24.0 g / 40.08 g/mol (molar mass of calcium)
Moles of calcium = 0.599 mol
2. Calculate the number of moles of nitrogen:
Moles of nitrogen = mass of nitrogen / molar mass of nitrogen
Moles of nitrogen = 12.0 g / 28.01 g/mol (molar mass of nitrogen)
Moles of nitrogen = 0.428 mol
Based on the stoichiometry of the balanced equation, the molar ratio between calcium and nitrogen is 3:1.
3. Determine the limiting reagent:
Since the ratio of calcium to nitrogen is less than 3:1, it means that calcium is present in excess and nitrogen is the limiting reagent.
Therefore, nitrogen is the limiting reagent in this reaction.
To determine the limiting reagent in a chemical reaction, you need to compare the number of moles of each reactant and see which one is present in the lesser amount.
Let's start by calculating the number of moles for each reactant:
The molar mass of calcium (Ca) is 40.1 g/mol.
Number of moles of calcium = mass of calcium / molar mass of calcium
Number of moles of calcium = 24.0 g / 40.1 g/mol
Number of moles of calcium = 0.598 mol
The molar mass of nitrogen (N2) is 28.0 g/mol.
Number of moles of nitrogen = mass of nitrogen / molar mass of nitrogen
Number of moles of nitrogen = 12.0 g / 28.0 g/mol
Number of moles of nitrogen = 0.429 mol
Now, let's examine the balanced chemical equation:
3Ca + N2 -> Ca3N2
According to the equation, the mole ratio between calcium and nitrogen is 3:1. This means that for every 3 moles of calcium, 1 mole of nitrogen is needed for complete reaction.
Now, let's determine which reactant is the limiting reagent:
To do this, we can compare the number of moles of nitrogen (0.429 mol) to the number of moles of calcium (0.598 mol) using the mole ratio from the balanced equation.
0.429 mol nitrogen * (3 mol calcium/1 mol nitrogen) = 1.288 mol calcium needed
Since the number of moles of nitrogen (0.429 mol) is less than the amount needed for complete reaction (1.288 mol), nitrogen is the limiting reagent.
Therefore, the limiting reagent is nitrogen (N2).