Ca+N-->Ca3N2
-What is limiting reagent?
-Mass of calcium nitride formed when 50.0g of calcium reacts with 50.0g of Nitrogen?
-Excess?
CAN SOMEONE DOUBLE CHECK MY ANSWERS:
Calcium is limiting reagent.
Mass of Ca3N2: 68.4g Ca3N2
Excess: 38.4N left over.
Ca = 40 g/mol
N = 14 g/mol
3 Ca = 120 g/mol of Ca3N2
2 N = 28 g/mol of Ca3N2
so we need 28 g N for every 120 g of Ca
so yes, if you have 50 g of each, then Ca is the limiting reagent
I only need 50 *28/120 = 21 grams of N so
50 -21 = 29 would be left over
Ca3N2 molar mass = 120 + 28 = 148 g/mol
148/120 * 50 = 61.7 g of Ca3N2 I get
Awesome that's what I got too.
Ah, it's time to clown around with some chemistry! Let's double-check your answers and have some fun while we're at it!
First off, you indeed identified calcium as the limiting reagent. Good job!
Now, let's calculate the mass of calcium nitride formed. We'll start by converting the given masses of calcium and nitrogen into moles:
Molar mass of calcium (Ca): 40.08 g/mol
Molar mass of nitrogen (N): 14.01 g/mol
Moles of calcium (Ca): 50.0 g / 40.08 g/mol = 1.25 mol
Moles of nitrogen (N): 50.0 g / 14.01 g/mol = 3.57 mol
The balanced equation tells us the mole ratio between calcium and calcium nitride is 3:1. Since calcium is our limiting reagent, we can only make as much calcium nitride as we have calcium. So, we have 1.25 moles of calcium, which should give us:
1.25 mol Ca * (1 mol Ca3N2 / 3 mol Ca) = 0.4167 mol Ca3N2
To find the mass of calcium nitride, we'll need to know its molar mass:
Molar mass of calcium nitride (Ca3N2): 20.08 g/mol (Ca) + 2 * 14.01 g/mol (N) = 68.10 g/mol
Mass of calcium nitride formed: 0.4167 mol Ca3N2 * 68.10 g/mol = 28.36 g Ca3N2
Great work so far! Now, let's check about the excess. Since calcium was the limiting reagent, the nitrogen will be in excess. To find the amount of nitrogen left over, we subtract what was consumed from the initial amount:
Initial moles of nitrogen (N): 3.57 mol
Consumed moles of nitrogen: 1.25 mol Ca * (1 mol N / 3 mol Ca) = 0.417 mol N
Excess moles of nitrogen: 3.57 mol - 0.417 mol = 3.153 mol N
To find the mass of nitrogen left over, we multiply the excess moles by the molar mass of nitrogen (N):
Mass of excess nitrogen: 3.153 mol N * 14.01 g/mol = 44.14 g N
So, your answer for the excess is slightly off. The correct mass of excess nitrogen left over is approximately 44.14 g.
Keep up the good work, and remember, chemistry can be clown-tastic fun!
To determine the limiting reagent, we need to compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation.
1. Begin by converting the masses of both calcium (Ca) and nitrogen (N2) to moles using their molar masses.
Molar mass of Ca = 40.08 g/mol
Molar mass of N2 = 28.02 g/mol
Moles of Ca = 50.0 g / 40.08 g/mol ≈ 1.25 mol
Moles of N2 = 50.0 g / 28.02 g/mol ≈ 1.78 mol
2. Next, apply the stoichiometric ratio for the balanced equation. From the equation Ca + N2 → Ca3N2, we can see that one mole of Ca reacts with one mole of N2 to produce one mole of Ca3N2.
To form 1.25 moles of Ca3N2, we need 1.25 moles of Ca and 1.25 moles of N2. However, we only have 1.25 moles of Ca, which means N2 is in excess.
3. To determine the mass of Ca3N2 formed, convert the moles of Ca3N2 to grams using its molar mass.
Molar mass of Ca3N2 = 68.04 g/mol
Mass of Ca3N2 = 1.25 mol × 68.04 g/mol ≈ 85.05 g
Therefore, the mass of calcium nitride formed when 50.0 g of calcium reacts with 50.0 g of nitrogen is approximately 85.05 g.
As for the excess, since we determined that nitrogen (N2) is in excess, we need to calculate the remaining amount of nitrogen after the reaction. The stoichiometric ratio tells us that one mole of nitrogen is required to react with one mole of calcium. Since we had 1.78 moles of N2 initially, and only 1.25 moles were used up, the excess amount is:
Excess N2 = 1.78 mol - 1.25 mol ≈ 0.53 mol
Finally, convert the moles of excess nitrogen to grams using its molar mass.
Mass of excess N2 = 0.53 mol × 28.02 g/mol ≈ 14.82 g
Therefore, the excess nitrogen left over is approximately 14.82 g.
To determine the limiting reagent, we need to compare the amount of reactant used to its stoichiometric coefficients in the balanced equation. In this case, the balanced equation is:
Ca + N2 → Ca3N2
To find the limiting reagent, we can convert the given masses of calcium and nitrogen to moles and then compare the mole ratio to see which reactant is in excess.
1. Calculating the moles of calcium:
Given mass of calcium = 50.0 g
Molar mass of calcium = 40.1 g/mol
Number of moles of calcium = mass / molar mass = 50.0 g / 40.1 g/mol = 1.25 mol
2. Calculating the moles of nitrogen:
Given mass of nitrogen = 50.0 g
Molar mass of nitrogen = 28.0 g/mol
Number of moles of nitrogen = mass / molar mass = 50.0 g / 28.0 g/mol = 1.79 mol
3. Comparing the mole ratio of calcium to nitrogen in the balanced equation:
From the balanced equation, we know that 3 moles of calcium react with 1 mole of nitrogen to form 1 mole of Ca3N2.
Mole ratio of calcium to nitrogen = 3:1
Since we have 1.25 moles of calcium and 1.79 moles of nitrogen, we can divide the number of moles of calcium by its stoichiometric coefficient (3) and the number of moles of nitrogen by its stoichiometric coefficient (1) to compare which reactant is in excess.
Calcium: 1.25 mol / 3 = 0.42
Nitrogen: 1.79 mol / 1 = 1.79
Since 0.42 < 1.79, it means that calcium is the limiting reagent. Therefore, the limiting reagent is calcium.
Now, let's calculate the mass of calcium nitride formed when 50.0 g of calcium reacts with 50.0 g of nitrogen.
1. Calculate the number of moles of calcium nitride formed:
According to the balanced equation, 1 mole of calcium reacts with 1 mole of nitrogen to form 1 mole of Ca3N2.
Since calcium is the limiting reagent, we use the moles of calcium to determine how many moles of Ca3N2 are formed.
Number of moles of Ca3N2 = 1.25 mol (moles of calcium)
2. Calculate the molar mass of Ca3N2:
Calcium nitride has a molar mass of 148.2 g/mol.
3. Calculate the mass of calcium nitride formed:
Mass of Ca3N2 = Number of moles of Ca3N2 x molar mass of Ca3N2
= 1.25 mol x 148.2 g/mol
= 185.25 g
Therefore, the mass of calcium nitride formed when 50.0 g of calcium reacts with 50.0 g of nitrogen is 185.25 g.
Regarding the excess, since calcium is the limiting reagent, there will be some nitrogen left over. To determine the amount of excess nitrogen, we can subtract the amount of nitrogen that reacted from the total amount of nitrogen initially present.
Initial moles of nitrogen: 1.79 mol
Moles of nitrogen used in the reaction: 1.25 mol (because it's the stoichiometric ratio with calcium)
Excess moles of nitrogen = Initial moles of nitrogen - Moles of nitrogen used in the reaction
= 1.79 mol - 1.25 mol
= 0.54 mol
To find the mass of the excess nitrogen, we multiply the excess moles by the molar mass of nitrogen:
Mass of excess nitrogen = Excess moles of nitrogen x molar mass of nitrogen
= 0.54 mol x 28.0 g/mol
= 15.1 g
So, the excess of nitrogen left over is 15.1 g.