If 23.5 g of ammonia (NH3) is dissolved in 1.0 L of solution, what is the molarity?

Phosphorous trichloride (PCl3) is produced

vladimir1956 [14]

The percentage yield obtained from the given reaction above is 74.8%

<h3>Balanced equation </h3>

P₄ + 6Cl₂ → 4PCl₃

Molar mass of P₄ = 31 × 4 = 124 g/mol

Mass of P₄ from the balanced equation = 1 × 124 = 124 g

Molar mass of PCl₃ = 31 + (35.5×3) = 137.5 g/mol

Mass of PCl₃ from the balanced equation = 4 × 137.5 = 550 g

<h3>SUMMARY</h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

<h3>How to determine the theoretical yield </h3>

From the balanced equation above,

124 g of P₄ reacted to produce 550 g of PCl₃

Therefore,

79.12 g of P₄ will react to produce = (79.12 × 550) / 124 = 350.9 g of PCl₃

<h3>How to determine the percentage yield </h3>

Actual yield of PCl₃ = 262.6 g
Theoretical yield of PCl₃ = 350.9 g

Percentage yield =?

Percentage yield = (Actual /Theoretical) × 100

Percentage yield = (262.6 / 350.9) × 100

Percentage yield = 74.8%

Learn more about stoichiometry:

brainly.com/question/14735801

4 0

2 years ago

If light is what is virtually the fastest thing ever then how can the dark keep up with it?

Sav [38]

Dark is just absence of light like if the re container a and container b u nee to pour hcl from a to so a has hcl while b has absence of hcl

tell me if u need more examples.hope it helps

5 0

4 years ago

Os This question is about elements from these families alkali metals, alkaline earth metals (Group

Lyrx [107]

Answer:

* is a soft,silvery metal that reacts violently with water

5 0

3 years ago

Read 2 more answers

The heats of combustion of ethane (C2H6) and butane (C4H10) are 52 kJ/g and 49 kJ/g, respectively. We need to produce 1.000 x 10

LekaFEV [45]

Answer :

(1) The number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 19.23 g and 20.41 g respectively.

(2) The number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ are 0.641 moles and 0.352 moles respectively.

(3) The balanced chemical equation for the combustion of the fuels.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

(4) The number of moles of $CO_2$ produced by burning each fuel is 1.28 mole and 1.41 mole respectively.

The fuel that emitting least amount of $CO_2$ is $C_2H_6$

Explanation :

First we have to calculate the number of grams needed of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

As, 52 kJ energy required amount of $C_2H_6$ = 1 g

So, 1000 kJ energy required amount of $C_2H_6$ = $\frac{1000}{52}=19.23g$

and,

As, 49 kJ energy required amount of $C_4H_{10}$ = 1 g

So, 1000 kJ energy required amount of $C_4H_{10}$ = $\frac{1000}{49}=20.41g$

Now we have to calculate the number of moles of each fuel $(C_2H_6)\text{ and }(C_4H_{10})$ .

Molar mass of $C_2H_6$ = 30 g/mole

Molar mass of $C_4H_{10}$ = 58 g/mole

$\text{ Moles of }C_2H_6=\frac{\text{ Mass of }C_2H_6}{\text{ Molar mass of }C_2H_6}=\frac{19.23g}{30g/mole}=0.641moles$

and,

$\text{ Moles of }C_4H_{10}=\frac{\text{ Mass of }C_4H_{10}}{\text{ Molar mass of }C_4H_{10}}=\frac{20.41g}{58g/mole}=0.352moles$

Now we have to write down the balanced chemical equation for the combustion of the fuels.

The balanced chemical reaction for combustion of $C_2H_6$ is:

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

and,

The balanced chemical reaction for combustion of $C_4H_{10}$ is:

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

Now we have to calculate the number of moles of $CO_2$ produced by burning each fuel to produce 1000 kJ.

$C_2H_6+\frac{7}{2}O_2\rightarrow 2CO_2+3H_2O$

From this we conclude that,

As, 1 mole of $C_2H_6$ react to produce 2 moles of $CO_2$

As, 0.641 mole of $C_2H_6$ react to produce $0.641\times 2=1.28$ moles of $CO_2$

and,

$C_4H_{10}+\frac{13}{2}O_2\rightarrow 4CO_2+5H_2O$

From this we conclude that,

As, 1 mole of $C_4H_{10}$ react to produce 4 moles of $CO_2$

As, 0.352 mole of $C_4H_{10}$ react to produce $0.352\times 4=1.41$ moles of $CO_2$

So, the fuel that emitting least amount of $CO_2$ is $C_2H_6$

5 0

3 years ago

Hibernation is _______. A. An example of an abiotic cycle b. Triggered by changes in the weather c. Caused only by biotic factor

natta225 [31]

The correct answer is option b, that is, triggered by changes in the weather.

Hibernation refers to a state of lower metabolism and inactivity usually witnessed in endotherms. This condition is featured by a slow heart rate, slow breathing, low metabolic rate, and low body temperature. It assists the majority of the endotherms to thrive in the cold seasons. For example, hibernation of a bear in winters. Therefore, weather change can stimulate hibernation.

6 0

4 years ago