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3 years ago

How many moles of ammonia (nh3) would be produced if 2.5 moles of nitrogen (n2) reacted with excess hydrogen (h2)?

1 answer:

6 0

Firstly, a balanced equation has to be written for the production of ammonia (NH₃) from hydrogen gas (H₂) and nitrogen gas (N₂):
N₂ + 3H₂ → 2NH₃

Now, the mole ratio of N₂ : NH₃ is 1 : 2 based on the coefficients of the balanced equation.

If the moles of N₂ = 2.5 moles

then the moles of NH₃ produced = 2.5 mol × 2
= 5 mol

Thus, the moles of ammonia produced when 2.5 mol of nitrogen gas is combined with excess hydrogen gas is 5 mol.

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What is the theoretical yield of aluminum oxide if 3.00 mol of aluminum metal is exposed to 2.55 mol of oxygen?

lisabon 2012 [21]

Theoretical yield of Al₂O₃: 1.50 mol.

<h3>Explanation</h3>

$2 \; \text{Al} + \dfrac{3}{2} \; \text{O}_2 \to {\bf 1} \; \text{Al}_2\text{O}_3$ ;

$4 \; \text{Al} + 3 \; \text{O}_2 \to 2 \; \text{Al}_2\text{O}_3 \; \textit{Balanced}$ .

How many moles of aluminum oxide formula units will be produced <em>if</em> aluminum is the limiting reactant?

Aluminum reacts to aluminum oxide at a two-to-one ratio.

$3.00 \times \dfrac{1}{2} = 1.50 \; \text{mol}$ .

As a result, 3.00 moles of aluminum will give rise to 1.50 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced <em>if</em> oxygen is the limiting reactant?

Oxygen reacts to produce aluminum oxide at a three-to-two ratio.

$2.55 \times \dfrac{2}{3} = 1.70 \; \text{mol}$

As a result, 2.55 moles of oxygen will give rise to 1.70 moles of aluminum oxide.

How many moles of aluminum oxide formula units will be produced?

Aluminum is the limiting reactant. Only 1.50 moles of aluminum oxide formula units will be produced. 1.70 moles isn't feasible since aluminum would run out by the time 1.50 moles was produced.

4 0

3 years ago

A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?

Licemer1 [7]

Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

$\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles$

Now we have to calculate the moles of $Br_2$

${\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles$

${\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles$

The balanced chemical reaction is,

$2Na(s)+Br_2(g)\rightarrow 2NaBr$

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = $\frac{2}{1}\times 0.189=0.378$ moles of Sodium bromide

Now we have to calculate the percent yield of reaction

$\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%$

Therefore, the percent yield is, 93.4%

3 0

3 years ago

What state does a maple tree enter during the winter?

Crank

D. dormancy is the correct answer

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3 years ago

Какая система называется моделью химия

OlgaM077 [116]

I don’t understand, what does that say in english?

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2 years ago

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How can the knowledge of a material’s chemical makeup or use be utilized?

shutvik [7]

To identify minerals

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3 years ago

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