Calcium reacts with nitrogen to form calcium nitride. Indicate which reactant is limiting, calculate the mass of calcium nitride formed when 50.0g of calcium reacts with 50.0g of Nitrogen. How much excess reactant do you have?
3Ca + N2 ==> Ca3N2
mols Ca = grams/atomic mass
mols N2 = grams/molar mass
Using the coefficients in the balanced equation, convert mols Ca to mols product.
Do the same for mols N2 to mols product.
It is likely that these two values for mols product will not agree which mans one of them is wrong. The correct value in limiting reagent problems is ALWAYS the smaller value and the reagent producing that number is the limiting reagent.
g Ca3N2 = mols Ca3N2 x molar mass Ca3N2.
How much of the excess reagent is left?
Using the coefficients in the balanced equation (just as you did in converting to mols product), convert mols of the limiting reagent to mols of the non-limiting reagent. This tells you how much of that reagent was used. Subtract amount used from initial amount to find amount not reacted.
Post your work if you get stuck.
To determine the limiting reactant and the amount of calcium nitride formed, we need to compare the amounts of calcium and nitrogen available.
1. Calculate the moles of calcium:
Given:
Mass of calcium = 50.0 g
Atomic mass of calcium (Ca) = 40.08 g/mol
Moles of calcium = mass of calcium / atomic mass of calcium
Moles of calcium = 50.0 g / 40.08 g/mol
Moles of calcium = 1.248 mol (rounded to three decimal places)
2. Calculate the moles of nitrogen:
Given:
Mass of nitrogen = 50.0 g
Atomic mass of nitrogen (N) = 14.01 g/mol
Moles of nitrogen = mass of nitrogen / atomic mass of nitrogen
Moles of nitrogen = 50.0 g / 14.01 g/mol
Moles of nitrogen = 3.568 mol (rounded to three decimal places)
3. Determine the ratio of moles:
The balanced equation for the reaction is:
3Ca + N₂ → Ca₃N₂
By examining the coefficients, we can see that the ratio of calcium to nitrogen is:
3 moles of calcium : 1 mole of nitrogen
4. Determine the limiting reactant:
To determine the limiting reactant, compare the moles of each reactant to the stoichiometric ratio.
Calcium: 1.248 mol
Nitrogen: 3.568 mol
Since the stoichiometric ratio is 3:1, we can see that there is more nitrogen than the stoichiometric ratio requires. Thus, the limiting reactant is calcium.
5. Calculate the mass of calcium nitride formed:
Since calcium is the limiting reactant, it will determine the maximum amount of calcium nitride formed.
By comparing the stoichiometric ratio, we can see that for every 3 moles of calcium, we get 1 mole of calcium nitride.
Moles of calcium nitride = (1.248 mol calcium / 3 mol calcium nitride) ≈ 0.416 mol calcium nitride
Mass of calcium nitride = moles of calcium nitride × molar mass of calcium nitride
Molar mass of calcium nitride (Ca₃N₂) = (3 × atomic mass of calcium) + (2 × atomic mass of nitrogen)
Molar mass of calcium nitride (Ca₃N₂) = (3 × 40.08 g/mol) + (2 × 14.01 g/mol)
Molar mass of calcium nitride (Ca₃N₂) = 120.24 g/mol + 28.02 g/mol
Molar mass of calcium nitride (Ca₃N₂) = 148.26 g/mol
Mass of calcium nitride = 0.416 mol × 148.26 g/mol
Mass of calcium nitride = 61.54 g (rounded to two decimal places)
6. Determine the excess reactant:
To find the excess reactant, we need to calculate the moles of nitrogen used in the reaction. Since calcium is the limiting reactant, it will react completely, so the excess is determined by the moles of nitrogen.
Moles of nitrogen used = (stochiometric ratio × moles of calcium) / moles of calcium nitride
Moles of nitrogen used = (1 mol nitrogen / 3 mol calcium) × 1.248 mol calcium / 0.416 mol calcium nitride
Moles of nitrogen used ≈ 1.997 mol nitrogen
Moles of nitrogen in excess = moles of nitrogen initially - moles of nitrogen used
Moles of nitrogen in excess = 3.568 mol - 1.997 mol
Moles of nitrogen in excess ≈ 1.571 mol (rounded to three decimal places)
Mass of nitrogen in excess = moles of nitrogen in excess × atomic mass of nitrogen
Mass of nitrogen in excess = 1.571 mol × 14.01 g/mol
Mass of nitrogen in excess ≈ 21.99 g (rounded to two decimal places)
Therefore, the mass of calcium nitride formed is approximately 61.54 g, and there is approximately 21.99 g of excess nitrogen remaining.
To determine which reactant is limiting, we need to compare the stoichiometric ratios of the reactants to the balanced chemical equation for the reaction between calcium and nitrogen, which is Ca + N₂ -> Ca₃N₂.
First, we need to find the number of moles of each reactant:
1. Calculate the number of moles of calcium by dividing the given mass (50.0g) by its molar mass (40.08 g/mol):
Moles of calcium = 50.0g / 40.08 g/mol = 1.25 mol
2. Calculate the number of moles of nitrogen by dividing the given mass (50.0g) by its molar mass (28.02 g/mol):
Moles of nitrogen = 50.0g / 28.02 g/mol = 1.784 mol
Now, let's examine the stoichiometric ratios of the reactants in the balanced equation:
Ca : N₂ : Ca₃N₂ = 1 : 1 : 1 (from the balanced chemical equation)
Comparing the number of moles of calcium (1.25 mol) to nitrogen (1.784 mol), we see that the ratio of calcium to nitrogen is 1:1, which means they are present in equal stoichiometric amounts.
Since they are present in equal quantities, neither reactant is limiting; therefore, all of the calcium and nitrogen will react completely.
To calculate the mass of calcium nitride formed, we can use the molar mass of calcium nitride (Ca₃N₂), which can be obtained by summing the molar masses of its constituent elements:
Molar mass of Ca₃N₂ = (3 × molar mass of Ca) + (2 × molar mass of N) = (3 × 40.08 g/mol) + (2 × 28.02 g/mol) = 146.26 g/mol
Now, we can calculate the mass of calcium nitride formed using the number of moles obtained earlier:
Mass of calcium nitride formed = Moles of calcium nitride × molar mass of Ca₃N₂
= 1.25 mol × 146.26 g/mol
= 182.825 g
Therefore, 182.825 grams of calcium nitride will be formed when 50.0 grams of calcium reacts with 50.0 grams of nitrogen.
Since neither reactant is limiting, there is no excess reactant.