6Li(s) + N2(g) 2Li3N(s)
a) What mass of lithium is required to produce 0.575 g of Li3N in the balanced reaction?
b) What mass of nitrogen is required for the reaction?
-------------------------------------------
I tried dimensional analysis in a couple different ways but I guess the numbers were wrong because I'm getting the wrong answer.
Here's the work:
a) .575 g Li3N(1 mol Li3N/34.83 g Li3N)(6 mol Li/2 mol Li3N)(6.941 g L/1 mol Li) = 0.334 g Li
Plug in the molar ratios and masses of N2 for part B.
To solve this problem, we need to use stoichiometry to convert between the given mass of Li3N and the mass of lithium and nitrogen required for the reaction.
First, let's start with part (a):
a) What mass of lithium is required to produce 0.575 g of Li3N in the balanced reaction?
From the balanced equation: 6Li(s) + N2(g) → 2Li3N(s), we can see that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
Step 1: Calculate the molar mass of Li3N:
The molar mass of Li3N = (3 × molar mass of Li) + (1 × molar mass of N)
Molar mass of Li3N = (3 × 6.94 g/mol) + (1 × 14.01 g/mol) = 50.73 g/mol
Step 2: Calculate the number of moles of Li3N:
Number of moles = mass / molar mass
Number of moles of Li3N = 0.575 g / 50.73 g/mol = 0.01133 mol
Step 3: Use the stoichiometric ratio to find the number of moles of Li:
From the balanced equation, we know that 6 moles of Li react to produce 2 moles of Li3N.
So, the ratio of moles of Li to moles of Li3N is 6:2 = 3:1.
Number of moles of Li = (3/1) × (0.01133 mol) = 0.03399 mol
Step 4: Calculate the mass of lithium needed:
Mass = moles × molar mass
Mass of lithium = 0.03399 mol × 6.94 g/mol = 0.2359 g
Therefore, the mass of lithium required to produce 0.575 g of Li3N is 0.2359 g.
Now, let's move on to part (b):
b) What mass of nitrogen is required for the reaction?
Again, using the stoichiometry from the balanced equation, we can see that 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
Step 1: Calculate the number of moles of Li3N:
Number of moles of Li3N = 0.01133 mol (from part a)
Step 2: Use the stoichiometric ratio to find the number of moles of N2:
From the balanced equation, we can see that the ratio of moles of N2 to moles of Li3N is 1:2.
Therefore, the number of moles of N2 = (1/2) × (0.01133 mol) = 0.005665 mol
Step 3: Calculate the mass of nitrogen needed:
Mass = moles × molar mass
Mass of nitrogen = 0.005665 mol × 28.02 g/mol = 0.1589 g
Therefore, the mass of nitrogen required for the reaction is 0.1589 g.
To solve these stoichiometry problems, we will use the balanced chemical equation and conversion factors.
Given the balanced equation:
6Li(s) + N2(g) → 2Li3N(s)
a) To find the mass of lithium required to produce 0.575g of Li3N, we can use the molar mass and stoichiometry.
1. Calculate the molar mass of Li3N.
Li: 3 lithium atoms × atomic mass of Li = 3 × (6.941 g/mol) = 20.823 g/mol
N: 1 nitrogen atom × atomic mass of N = 1 × (14.007 g/mol) = 14.007 g/mol
Total molar mass of Li3N = 20.823 g/mol + 14.007 g/mol = 34.830 g/mol
2. Use the stoichiometry of the balanced equation to convert grams of Li3N to moles of Li3N.
Moles of Li3N = Grams of Li3N / Molar mass of Li3N
= 0.575 g / 34.830 g/mol
≈ 0.0165 mol
3. Use the stoichiometry of the balanced equation to convert moles of Li3N to moles of Li.
From the balanced equation, 2 moles of Li3N are formed from 6 moles of Li.
Moles of Li = (0.0165 mol Li3N) × (6 mol Li / 2 mol Li3N)
= 0.0495 mol Li
4. Finally, calculate the mass of lithium required.
Mass of Li = Moles of Li × Molar mass of Li
= 0.0495 mol × 6.941 g/mol
≈ 0.3439 g
Therefore, approximately 0.3439 grams of lithium are required to produce 0.575 grams of Li3N.
b) To find the mass of nitrogen required for the reaction, we follow a similar process.
1. Calculate the molar mass of N2.
Molar mass of N2 = 2 nitrogen atoms × atomic mass of N = 2 × (14.007 g/mol) = 28.014 g/mol
2. Use the stoichiometry of the balanced equation to convert moles of Li3N to moles of N2.
From the balanced equation, 6 moles of Li react with 1 mole of N2 to produce 2 moles of Li3N.
Moles of N2 = (0.0165 mol Li3N) × (1 mol N2 / 2 mol Li3N)
= 0.00825 mol N2
3. Finally, calculate the mass of nitrogen required.
Mass of N2 = Moles of N2 × Molar mass of N2
= 0.00825 mol × 28.014 g/mol
≈ 0.2312 g
Therefore, approximately 0.2312 grams of nitrogen are required for the reaction.
Make sure to double-check your calculations and unit conversions to ensure accurate results.
If you had shown your work instead of "but I guess the numbers were wrong" we could tell instantly what was wrong.
a.
mols Li3N = grams/molar mass = ?
Then ?mol Li3N x (6 mol Li/2 mol Li3N) = ?
grams Li = mols Li x atomic mass Li = ?
Also I suggest you find th arrow key on your computer and use it. --> or ==>