Describe how naturally acidic rainwater can affect a mountain of limestone.

If 8.50 g of phosphorus reacts with hydrogen gas at 2.00 atm in a 10.0-L container at 298 K, calculate the moles of PH3 produced

ahrayia [7]

Answer:

The moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

Explanation:

Phosphorus reacts with H₂ according to the balanced equation:

P₄ (s) + 6 H₂ (g) ⇒ 4 PH₃ (g)

By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of each compound participate in the reaction:

P₄: 1 mole
H₂: 6 moles
PH₃:4 moles

Being the molar mass of the compounds:

P₄: 124 g/mole
H₂: 2 g/mole
PH₃: 34 g/mole

The following mass amounts of each compound participate in the reaction:

P₄: 1 mole* 124 g/mole= 124 g
H₂: 6 mole* 2 g/mole= 12 g
PH₃: 4 moles* 34 g/mole= 136 g

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

In this case you know:

P= 2 atm
V= 10 L
n= ?
R= 0.082 $\frac{atm*L}{mol*K}$
T= 298 K

Replacing:

2 atm*10 L= n*0.082 $\frac{atm*L}{mol*K}$ *298 K

and solving you get:

$n=\frac{2 atm*10 L}{0.082\frac{atm*L}{mol*K}*298 K }$

n=0.818 moles

The limiting reagent is one that is consumed first in its entirety, determining the amount of product in the reaction. When the limiting reagent is finished, the chemical reaction will stop.

To determine the limiting reagent, you can use a simple rule of three as follows: if 6 moles of H₂ react with 124 g of P₄, 0.818 moles of H₂ with how much mass of P₄ will it react?

$mass of P_{4}=\frac{0.818 moles of H_{2}*124 grams of P_{4}}{6 moles of H_{2}}$

mass of P₄= 16.90 grams

But 16.90 grams of P₄ are not available, 8.50 grams are available. Since you have less mass than you need to react with 0.818 moles of H₂, phosphorus P₄ will be the limiting reagent.

Then you can apply the following rules of three:

If 124 grams of P₄ produce 4 moles of PH₃, 8.50 grams of P₄, how many moles do they produce?

$moles of PH_{3} =\frac{8.5 grams of P_{4}*4 moles of PH_{3} }{124grams of P_{4}}$

moles of PH₃=0.2742

If 124 grams of P₄ react with 6 moles of H₂, 8.50 grams of P₄ with how many moles of H₂ do they react?

$moles of H_{2} =\frac{8.5 grams of P_{4}*6 moles of H_{2} }{124grams of P_{4}}$

moles of H₂= 0.4113

If you have 0.818 moles of H₂, the number of moles of gas H₂ present at the end of the reaction is calculated as:

0.818 - 0.4113= 0.4067

Then the total number of moles of gas present at the end of the reaction will be the sum of the moles of PH₃ gas and H₂ gas that did not react:

0.2742 + 0.4067= 0.6809

Finally, the moles of PH₃ produced are 0.2742 and the total number of moles of gas present at the end of the reaction is 0.6809.

5 0

3 years ago

the development and use of the ______ a common and important piece of laboratory equipment, led scientist to the discovery of se

Liula [17]

The answer is the Microscope.

6 0

4 years ago

At the freezing point, the particles in an object have no kinetic energy. true or false?

Sliva [168]

False because particles stop moving or study slow down.

5 0

3 years ago

Read 2 more answers

How many of the following statements about silver acetate, AgCH3COO, are true? i) More AgCH3CoO(S) will dissolve if the pH of th

shutvik [7]

Answer:

All three statements are true

Explanation:

Solubility equilibrium of silver acetate:

$AgCH_{3}COO\rightleftharpoons Ag^{+}+CH_{3}COO^{-}$

If pH is increased then concentration of $H^{+}$ increases in solution resulting removal of $CH_{3}COO^{-}$ by forming $CH_{3}COOH$ . Hence, according to Le-chatelier principle, equilibrium will shift towards right. So more $AgCH_{3}COO$ will dissolve
If $AgNO_{3}$ is added then concentration of $Ag^{+}$ increases in solution resulting shifting of equilibrium towards left in accordance with Le-chatelier principle. So less $AgCH_{3}COO$ will dissolve
Insoluble precipitate of AgOH is formed by adding NaOH in solution resulting removal of $Ag^{+}$ . So, more $AgCH_{3}COO$ will dissolve

Hence all three statements are true

8 0

3 years ago

A student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in ea

Natali [406]

Explanation:

the experiment conducted is the student adds sugar to a cup of iced tea and a cup of hot tea. She notices that the time needed for the sugar to dissolve in each cup is different. She thinks this has something to do with the temperature of the tea

hypothesis: If the student puts the sugar in both glasses of tea, then the sugar in the hot tea should dissolve quicker.

4 0

3 years ago