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damaskus [11]
3 years ago
10

with an excess of oxygen gas according to the following chemical equation. 4 P 502 > 2 P2Os What mass of P2O5 can be produced

when a 172.1 g sample of phosphorus reacts
Chemistry
1 answer:
MArishka [77]3 years ago
4 0

Answer:

394.76g

Explanation:

Given parameters:

Mass of Phosphorus = 172.1g

Unknown:

Mass of P₂O₅ = ?

Solution:

We should work from the known specie to the unknown in this problem. The limiting reactant is the reactant in short supply and it determines the extent of the reaction. This is the given phosphorus.

Oxygen is in the excess.

The balanced reaction equation:

            4P + 5O₂   →   2P₂O₅

let us find the number of moles of phosphorus first,

 Number of moles of phosphorus = \frac{mass}{molar mass}

Molar mass of phosphorus = 31g/mole

Number of moles = \frac{172.1}{31}  = 5.55mole

   4 moles of phosphorus gives 2 moles of P₂O₅

 5.55 mole of P will produce \frac{5.55 x 2}{4}   = 2.78moles

Mass of P₂O₅  = number of moles x molar mass

 Molar mass of P₂O₅  = (31 x 2)  +  (16 x 5)  = 142g/mole

Mass of P₂O₅  = 2.78moles x 142g/mole = 394.76g

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1. If you have a sample of gas at a pressure of 16 atm, what will the pressure be if the volume is halved?
vagabundo [1.1K]

Answer:

1. The pressure will be 32 atm, twice the initial pressure.

2. The pressure will be 1.83 atm, one third of the initial pressure.

Explanation:

Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at a constant temperature.

This law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure." This means that if the pressure increases, the volume decreases, while if the pressure decreases, the volume increases.

Boyle's law is expressed mathematically as:

Pressure * Volume = constant

or P * V = k

Ahora es posible suponer que tienes un cierto volumen de gas V1 que se encuentra a una presión P1 al comienzo del experimento. Si varias el volumen de gas hasta un nuevo valor V2, entonces la presión cambiará a P2, y se cumplirá:

P1*V1=P2*V2

1. In this case:

  • P1= 16 atm
  • V1
  • P2= ?
  • V2= V1÷2= \frac{V1}{2} because the volume is halved.

So:

16 atm*V1= P2* \frac{V1}{2}

Solving:

\frac{16 atm*V1*2}{V1}=P2

16 atm*2= P2

32 atm= P2

<u><em>The pressure will be 32 atm, twice the initial pressure.</em></u>

2. Now

  • P1= 5.5 atm
  • V1
  • P2= ?
  • V2= V1*3 because the volume is tripled.

So:

5.5 atm*V1= P2* V1*3

Solving:

\frac{5.5 atm*V1}{3*V1}=P2

\frac{5.5 atm}{3}= P2

1.83 atm= P2

<u><em>The pressure will be 1.83 atm, one third of the initial pressure.</em></u>

7 0
3 years ago
Certain compound contains 7.3% carbon, 4.5% hydrogen, 36.4% oxygen, and 31.8% nitrogen. It’s reality molecular mass is 176.0. Fi
erastovalidia [21]

Answer:

The answer to your question is:

Explanation:

Data

carbon        7.3%          =     7.3g

hydrogen    4.5%         =      4.5g

oxygen       36.4%         =     36.4 g

nitrogen     31.8%         =     31.8 g

Now

For carbon

                    12 g --------------------1 mol

                    7.3 g     -------------     x

                       x = 7.3/12 = 0.608 mol

For hydrogen

                 1 g   --------------------  1 mol

                 4.5 g  ------------------    x

                   x = 4.5 mol

For oxygen

             16 g ------------------- 1 mol

             36.4 g ----------------    x

             x = 2.28 mol

For nitrogen

              14 g   ----------------   1 mol

              31.8 g ---------------    x

             x = 2.27 mol

Now divide by the lowest result, the is 0.608 from carbon

carbon              0.608/0.608 = 1

hydrogen           4.5/ 0.608 = 7.4

oxygen              2.28/0.608 = 3.75

nitrogen             2.27/0.608 = 3.73

Empirical formula = CH₇O₄N₄

     

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