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Pavlova-9 [17]
3 years ago
12

The Haber process is used to make ammonia, which is an important source of nitrogen that can be metabolized by plants. Using the

equation below, determine how many kilograms of ammonia will be formed from 89.5 kg of hydrogen. N2(g) + 3H2(g) → 2NH3(g)
Chemistry
1 answer:
Dimas [21]3 years ago
7 0

Answer:

506kg

Explanation:

From the reaction equation, we can see that 3 moles of hydrogen yielded 2 moles of ammonia.

Now we need to calculate the number of moles of hydrogen reacted. This mathematically equals the mass of hydrogen reacted divided by the molecular mass of the hydrogen molecule.

The mass of hydrogen reacted is 89.5kg. This equals 89.5 × 1000g which equals 89500g. The molecular mass of hydrogen is 2g/mole ( 1 atom of hydrogen has an atomic mass unit of 1 and 1 molecule of hydrogen has 2 atoms).

Now we mathematically calculate the number of moles of hydrogen reacted to be 89500/2 which equals 44,750 moles.

If 3 moles of hydrogen yielded 2 moles of ammonia gas, then 44, 750 moles will yield 44,750 × (2/3) which equals 29,833 moles of Ammonia gas.

Now, let's calculate the mass of Ammonia gas yielded. This mathematically equals the number of moles of ammonia gas yielded multiplied by the molar mass of ammonia gas. The molar mass of ammonia gas is (14 + 3(1)) which equals 17g/mole.

Hence, the mass yielded equals 29,833 × 17 = 506167g

This equals 506,167 ÷ 1000 = 506.167kg.

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The total percent yield:

After the combustion reaction with methane, the percent yield was 66.7%.  

Combustion of Methane:

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  • Methane burns in the presence of enough oxygen to produce carbon dioxide (CO₂) and water (H₂O).
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The other reactant, air's excess oxygen, is always present, making methane the limiting reactant. As a result, the amount of CH₄ burned will determine how much CO₂ and H₂O are produced.

The following chemical process produces carbon dioxide from methane:

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Calculations:

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All calculations will be based on the amount of methane because the problem specifies that it is the limiting reagent:

12.0g of CH₄ × (1 mol of CH₄/16g CH₄) × (1 mole of CO₂/1 mole of CH₄) × (44g CO₂/1 mole of CO₂)

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<h3>Further explanation</h3>

Given

78% Nitrogen by volume

Required

The solubility of nitrogen in water

Solution

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Can be formulated  

S = kH. P.  

S = gas solubility, mol / L  

kH = Henry constant, mol / L.atm  

P = partial gas pressure  

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<span>35.0 mL of 0.210 M
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