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zimovet [89]
3 years ago
12

Consider the following reaction: 2 NO(g) + 5 H2(g) → 2 NH3(g) + 2 H2O(g) Which set of solution maps would be needed to calculate

the maximum amount of ammonia (NH3), in grams, that can be synthesized from 45.8 g of nitrogen monoxide (NO) and 12.4 g of hydrogen (H2)? I. g NO → mol NO → mol NH3 → g NH3 II. g H2 → mol H2 → mol NH3 → g NH3 III. g NO → mol NO → mol H2O → g H2O IV. g H2 → mol H2 → mol H2O → g H2O
Chemistry
1 answer:
Leto [7]3 years ago
7 0

Answer : The correct option is, (I) gNO\rightarrow molNO\rightarrow molNH_3\rightarrow gNH_3

Solution : Given,

Mass of NO = 45.8 g

Mass of H_2 = 12.4 g

Molar mass of NO = 30 g/mole

Molar mass of H_2 = 2 g/mole

Molar mass of NH_3 = 17 g/mole

First we have to calculate the moles of NO and O_2.

\text{ Moles of }NO=\frac{\text{ Mass of }NO}{\text{ Molar mass of }NO}=\frac{45.8g}{30g/mole}=1.53moles

\text{ Moles of }H_2=\frac{\text{ Mass of }H_2}{\text{ Molar mass of }H_2}=\frac{12.4g}{2g/mole}=6.20moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2NO(g)+5H_2(g)\rightarrow 2NH_3(g)+2H_2O(g)

From the balanced reaction we conclude that

As, 2 mole of NO react with 5 mole of H_2

So, 1.53 moles of NO react with \frac{1.53}{2}\times 5=3.82 moles of H_2

From this we conclude that, H_2 is an excess reagent because the given moles are greater than the required moles and NO is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of NH_3

From the reaction, we conclude that

As, 2 mole of NO react to give 2 mole of NH_3

So, 1.53 mole of NO react to give 1.53 mole of NH_3

Now we have to calculate the mass of NH_3

\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3

\text{ Mass of }NH_3=(1.53moles)\times (17g/mole)=26.0g

Therefore, the maximum mass of NH_3 produced 26.0 grams.

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The elements occurring in nature exist as multiple isotopes.

When we take into account the existence of these isotopes and their relative abundance (percent), the average atomic mass of that element can be computed, which is given by the following formula,

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The mass of Indium-113 is 112.90 u.

The mass of Indium-115 is 114.90 u.

The average atomic mass of Indium is 114.82 u.

Let the %age of isotope 1(Indium-113) be X.

Then, the %age of isotope 2(Indium-115) would be 100-X.

Applying the values in the formula,

Average atomic mass = 112.90X+114.90(100-X)

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On solving the above equation, the value of X comes out to be 4%.

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To know more about "Average Atomic Mass", refer to the following link:

brainly.com/question/13753702?referrer=searchResults

#SPJ4

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What is the ground state electron configuration for magnesium?
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