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Andre45 [30]
4 years ago
6

An industrial chemist is studying the rate of the haber synthesis: n2 + 3h2 → 2nh3. starting with a closed reactor containing 1.

35 mol/l of n2 and 0.35 mol/l of h2, she finds that the h2 concentration has fallen to 0.12 mol/l after 50. seconds. what is the average rate of reaction of h2 over this time? (enter in mol/liter/sec). what is the average rate of nh3 production for this example? (enter in mol/liter/sec)
Chemistry
2 answers:
Mila [183]4 years ago
6 0
Average  rate  of  reaction of  H2  over  this  time  is  calculated  as

initial  concentration  of  H2  minus  final  concentration  of  H2

That  is
{(0.35mol/l -0.12mol/l) / 50  sec}= 4.6 x 10 ^-3  mol/l/sec
 
Paul [167]4 years ago
5 0

1) Balanced chemical reaction: 3H₂ + N₂ → 2NH₃.

c₁(H₂) = 0.35 mol/L; initial concentration of hydrogen gas.

c₂(H₂) = 0.12 mol/L; concentration of hydrogen after 50 seconds.

Δt = 50 s.

the average rate of reaction = 1/3 · (Δc(H₂)/Δt).

the average rate of reaction of hydrogen = 1/3 · ((c₂(H₂)-c₁(H₂))/Δt).

the average rate of reaction = 1/3 · (0.12 mol/L - 0.35 mol/L) / 50 s.

the average rate of reaction of hydrogen = -0.00153 mol/Ls, it is negative because it measures the rate of disappearance of the reactants.

2) From chemical reaction: n(H₂) : n(NH₃) = 3 : 2.

c(NH₃) = 2 · 0.23 mol/L / 3.

c(NH₃) = 0.153 mol/L.

the average rate of reaction of ammonia = 1/2 · (0.153 / 50 s).

the average rate of reaction of ammonia = 0.00153 mol/L·s.

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ZanzabumX [31]

Answer:

2.95 g of CH₄

Explanation:

To start this, we determine the equation:

4H₂  + CO₂ →  CH₄  +  2H₂O

4 moles of hydrogen react to 1 mol of carbon dioxide in order to produce 1 mol of methane and 2 moles of water.

To determine the limiting reactant, we need to know the moles of each reactant.

8.1 g . 1 mol/ 44g = 0.184 moles of carbon dioxide

2.3 g . 1mol / 2g = 1.15 moles of hydrogen

4 moles of hydrogen react to 1 mol of CO₂

Then, 1.15 moles may react to (1.15 . 1) /4 = 0.2875 moles

We only have 0.184 moles of CO₂, so this is the limiting reactant. Not enough CO₂ to complete the 0.2875 moles that are needed.

Ratio is 1:1. 1 mol of CO₂ produces 1 mol of methane

Then, 0.184 moles of CO₂ will produce 0.184 moles of CH₄

We convert moles to mass: 0.184 mol . 16 g /mol = 2.95 g

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