In the reaction between 15.8 g of NH3 and 22.7 g oxygen, water and nitrogen monoxide are

Using stoichiometry, you predict that you should be able to use 314.0 g of Al to produce 1551 g of AlCi3. In your lab

Alex787 [66]

Answer:

90.26%

Explanation:

From the question given above, the following data were obtained:

Theoretical yield of AlCl₃ = 1551 g

Actual yield of AlCl₃ = 1400 g

Percentage yield =?

The percentage yield of the reaction can be obtained as follow:

Percentage yield = Actual yield / Theoretical yield × 100

Percentage yield = 1400 / 1551 × 100

Percentage yield = 140000 / 1551

Percentage yield = 90.26%

Thus, the percentage yield of the reaction is 90.26%

6 0

3 years ago

How many Noble gases are there?

melomori [17]

Answer:

six noble gases

Here are five of the six noble gases: helium, neon, argon, kypton and xeon. They're all colourless and transparent. Krypton and xeon form compounds only with difficulty. Helium, neon and argon don't form compounds at all.

3 0

3 years ago

Read 2 more answers

Calculate the speed of a wave with a wavelength of 6 meters and a frequency of 3 hertz.

Kruka [31]

The answer is 18 you multiply wavelength (6) times frequency (3)

3 0

3 years ago

Read 2 more answers

How many molecules of O2 are in 153.9 g O2?

MrRissso [65]

Answer:

2.895*10^24

Explanation:

mass of Oxygen give = 153.9g

molar mass of O2 molecule = 16*2=32g/mol

n= mole

To find the mole

n= mass/ molar mass

n= 153.9/32

n=4.81mol.

To find the number of molecules of o

Nm= number of molecule

Nn = Number of mole

NA = number of Avogados

Nm= Nn * NA

Nm= 4.81 *6.02*10^23

Nm= 2.895*10^24

3 0

3 years ago

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$

From the reaction, we conclude that

As, 1 mole of $NaCl$ react to give 1 mole of $NaHCO_3$

So, 0.243 moles of $NaCl$ react to give 0.243 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0

4 years ago