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Leviafan [203]
3 years ago
15

Benzoic acid, c6h5cooh, has a ka = 6.4 x 10-5. what is the concentration of h3o+ in a 0.5 m solution of benzoic acid? 3.2 x 10-5

m
Chemistry
1 answer:
FinnZ [79.3K]3 years ago
3 0
The answer for this issue is: 
The chemical equation is: HBz + H2O <- - > H3O+ + Bz- 
Ka = 6.4X10^-5 = [H3O+][Bz-]/[HBz] 
Let x = [H3O+] = [Bz-], and [HBz] = 0.5 - x. 
Accept that x is little contrasted with 0.5 M. At that point, 
Ka = 6.4X10^-5 = x^2/0.5 
x = [H3O+] = 5.6X10^-3 M 
pH = 2.25
(x is without a doubt little contrasted with 0.5, so the presumption above was OK to make) 
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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:

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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)

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