Answer:
Nitrogen is limiting reactant while hydrogen is in excess.
Explanation:
Given data:
Mass of N₂ = 25 g
Mass of H₂ = 25 g
Mass of ammonia formed = ?
Solution:
Chemical equation:
N₂ + 3H₂ → 2NH₃
Number of moles of Nitrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 28 g/mol
Number of moles = 0.89 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 25 g/ 2 g/mol
Number of moles = 12.5 mol
Now we will compare the moles of both reactant with ammonia.
H₂ ; NH₃
3 : 2
12.5 : 2/3×12.5 = 8.3
N₂ ; NH₃
1 : 2
0.89 : 2×0.89 = 1.78
The number of moles of ammonia produced by nitrogen are less thus nitrogen is limiting reactant while hydrogen is in excess.
To get a result with the best degree of precision, the
number of significant figures should be equal to the smallest number of
significant figures of the given numbers. In this case, the smallest is 3 as
given by the number 9.03 mL.
Therefore density is:
<span>11.50 g / 9.03 mL = 1.27 g/mL</span>
In general, we have this rate law express.:
![\mathrm{Rate} = k \cdot [A]^x [B]^y](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5Ey)
we need to find x and y
ignore the given overall chemical reaction equation as we only preduct rate law from mechanism (not given to us).
then we go to compare two experiments in which only one concentration is changed
compare experiments 1 and 4 to find the effect of changing [B]
divide the larger [B] (experiment 4) by the smaller [B] (experiment 1) and call it Δ[B]
Δ[B]= 0.3 / 0.1 = 3
now divide experiment 4 by experient 1 for the given reaction rates, calling it ΔRate:
ΔRate = 1.7 × 10⁻⁵ / 5.5 × 10⁻⁶ = 34/11 = 3.090909...
solve for y in the equation
![\Delta \mathrm{Rate} = \Delta [B]^y](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BB%5D%5Ey)

To this point,
![\mathrm{Rate} = k \cdot [A]^x [B]^1](https://tex.z-dn.net/?f=%5Cmathrm%7BRate%7D%20%3D%20k%20%5Ccdot%20%5BA%5D%5Ex%20%5BB%5D%5E1%20)
do the same to find x.
choose two experiments in which only the concentration of B is unchanged:
Dividing experiment 3 by experiment 2:
Δ[A] = 0.4 / 0.2 = 2
ΔRate = 8.8 × 10⁻⁵ / 2.2 × 10⁻⁵ = 4
solve for x for
![\Delta \mathrm{Rate} = \Delta [A]^x](https://tex.z-dn.net/?f=%5CDelta%20%5Cmathrm%7BRate%7D%20%3D%20%5CDelta%20%5BA%5D%5Ex)

the rate law is
Rate = k·[A]²[B]
Fe
Cobalt
Sodium
Sn
Phosphorus
Florine
I don’t know
Mg
I don’t know
Calcium
C
Pb
Silver
Zinc
Nickel