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Nutka1998 [239]
3 years ago
14

The first step in the ostwald process for producing nitric acid is as follows: 4nh3(g) + 5o2(g) ? 4no(g) + 6h2o(g). if the react

ion of 15.0 g of ammonia with 15.0 g of oxygen gas yields 8.70 g of nitric oxide, what is the percent yield of this reaction?
Chemistry
1 answer:
nevsk [136]3 years ago
4 0
4 moles of NH3 reacts with 5 moles of O2.
4 moles of NH3 = 4 (14 + 3*1) = 68g
5 moles of O2 = 5*32 = 160g
Therefore, 68g of NH3 reacts with 160g of O2
Mass of NH3 to react with 15g of oxygen = 15*68 / 160 = 6.375g, which is less than 15 g.
Mass of O2 to react with 15g of NH3 = 15*160 / 68 = 35.29g, which is more than 15g. 
Therefore, the limiting reactant is O2, which means the reaction will proceed until the 15 g of oxygen finish.

4 moles of NO = 4 (14+16) 120g
Therefore, 160g of O2 gives 120g of NO.
15g of O2 will give Xg of NO.
X (theoritical yeild) = 15*120 / 160 = 11.25g
Practical yield = 8.70g
Yield = practical yeild / theoritical yeild *100 = 8.70/11.25 = 77.33%
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3 years ago
When the concentration of A in the reaction A ..... B was changed from 1.20 M to 0.60 M, the half-life increased from 2.0 min to
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Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

Half-life

{t_{1/2}}A=2\ \text{min}

{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

The order of the reaction is 2.

t_{1/2}=\dfrac{1}{k[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{t_{1/2}[A]_0^{n-1}}\\\Rightarrow k=\dfrac{1}{2\times 1.2^{2-1}}\\\Rightarrow k=0.4167\ \text{M}^{-1}\text{min}^{-1}

The rate constant is 0.4167\ \text{M}^{-1}\text{min}^{-1}

3 0
3 years ago
Differentiate benzene and cyclohexane
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Benzene is an aromatic compound but cyclohexane is not aromatic.

Benzene is an unsaturated molecule, but cyclohexane is saturated.
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Carbon atoms in the benzene ring have sp2 hybridization where carbon atoms in the cyclohexane have sp3 hybridization.</span>
 
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<em />

<em>Explanation:</em>

<h3><em /></h3>
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Answer:

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