theoretical yield of ammonia (NH₃) = 121.38 g
The limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).
Explanation:
We have the following chemical reaction in which nitrogen react with hydrogen to produce ammonia:
N₂ + 3 H₂ → 2 NH₃
Now we need to calculate the number of moles of each reactant:
number of moles = mass / molar weight
number of moles of N₂ = 100 / 28 = 3.57 moles
number of moles of H₂ = 100 / 2 = 50 moles
We see from the chemical reaction that 3 moles of H₂ react with 1 mole of N₂ so 50 moles of H₂ react with 16.67 moles of N₂ which is way more than the available N₂ quantity of 3.57 moles, so the limiting reactant is nitrogen (N₂) and the excess reactant is hydrogen (H₂).
Knowing this we devise the following reasoning:
if 1 mole of N₂ produces 2 moles of NH₃
then 3.57 moles of N₂ produces X moles of NH₃
X = (3.57 × 2) / 1 = 7.14 moles of NH₃
mass = number of moles × molar weight
mass of NH₃ = 7.14 × 17 = 121.38 g (theoretical yield)
Learn more about:
limiting reactant
brainly.com/question/14111505
#learnwithBrainly
#6).
Every 1,000 mL makes 1 L
How many 1,000mL are there in 2,800 mL ?
That's division.
(2,800 mL) / (1,000 mL) = <em>2.8 L</em>
#7).
The 'perimeter' means the 'distance all the way around'.
You have to know that both sides of a rectangle are the same length,
and also the top and bottom are the same length.
So the perimeter of this rectangle is
(2 yd) + (4.5 yd) + (2.yd) + (4.5 yd) = 13 yd .
Oops. The problem wants to know the perimeter in feet.
So you have to know that each yard is the same as 3 feet.
In order to find the number of feet in 13 yards, you have to
take 3 feet 13 <em><u>times</u></em> .
(3 feet) times (13) = <em>39 feet .</em>
#8).
For this one, you have to know that every 36 inches makes 1 yard.
How many 36 inches are there in 48 inches ?
That's division.
(48 inches) / (36inches) = <em>1 and 1/2 yards</em> .
#9).
For this problem, you have to know how to handle a mixed number,
and you also have to know that there are 16 ounces in 1 pound.
Add up the fruit:
(3-1/2 pounds) + (4 pounds) + 2 pounds) = <em><u>9-1/2 pounds</u></em>
Now, remember that each pound is the same as 16 ounces. So if you
want to find the number of ounces in 9-1/2 pounds, you have to take
16 ounces 9-1/2 times .
(16 ounces) times (9-1/2) = <em>152 ounces</em>.
___________________________________
#10).
This one is just adding up some numbers. But after you finish doing that, you have to know that 1,000 meters is called '1 kilometer' .
Add up the distances that Omar ran:
(1,000 meters) + (1,625 meters) + (1,500 meters) = <em><u>4,125 meters</u></em>
The problem wants to know how many kilometers this is, so you have to figure out how many '1,000 meters' fit into 4,125 meters.
That's division.
(4,125 meters) / (1,000 meters) = <em>4.125 kilometers</em>
Answer:
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Explanation:
for the reaction
PCl₅(g) → PCl₃(g) + Cl₂(g)
where
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M
and [A] denote concentrations of A
if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M
therefore
Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M
[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M
[PCl₅] = 3.64*10⁻³ M
the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M
Answer:
See explanation
Explanation:
Given Einstein's theory of relativity, we have that;
E= mc^2
m= mass of the substance
c= speed of light
For one gram of the substance,
E= 1 ×10^-3 × (3 × 10^8)^2
E = 9 × 10^13 J
For 8.7 g of matter;
E = 8.7 × 10^-3× (3 × 10^8)
E= 7.83 ×10^ 14 J
Explanation:
Mass of fructose = 33.56 g
Mass of water = 18.88 g
Total mass of the solution = Mass of fructose + Mass of water = M
M = 33.56 g + 18.88 g =52.44 g
Volume of the solution = V = 40.00 mL
Density =
a) Density of the solution:
b) Molar mass of fructose = 180.16 g/mol
Moles of fructose = 
Molar mass of water = 18.02 g/mol
Moles of water= 
Mole fraction of fructose in this solution:
Mole fraction of water = 
c) Average molar mass of of the solution:
=
d) Mass of 1 mole of solution = 42.50 g/mol
Density of the solution = 1.311 g/mL
d) Specific molar volume of the solution:
