What is a photon?

Plz help guys ASAP! Thanks in advance

MariettaO [177]

Answer:

3.6 moles

Explanation:

The balanced equation for the reaction is given below:

N₂ + 3H₂ —> 2NH₃

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂ to produce 2 moles NH₃.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

1 mole of N₂ reacted with 3 moles of H₂.

Therefore, 3.2 moles of N₂ will react with = 3.2 × 3 = 9.6 moles of H₂.

From the calculation made above, we can see that it will take a higher amount (i.e 9.6 moles) of H₂ than what was given (i.e 5.4 moles) to react completely with 3.2 moles of N₂.

Therefore, H₂ is the limiting reactant and N₂ is the excess reactant.

Finally, we the greatest quantity of ammonia, NH₃ produced from the reaction.

In this case, the limiting reactant will be use because all of it is consumed in the reaction.

The limiting reactant is H₂ and greatest quantity of ammonia, NH₃ produced can be obtained as follow:

From the balanced equation above,

3 moles of H₂ reacted to produce 2 moles NH₃.

Therefore, 5.4 of H₂ will react to produce = (5.4 × 2)/3 = 3.6 moles of NH₃

Thus, the greatest quantity of ammonia, NH₃ produced from the reaction is 3.6 moles

4 0

3 years ago

What volume of nitrogen dioxide is formed at 735 torr and 28.2 °C by reacting 3.56 cm3 of copper (d = 8.95 g/cm3) with 200 mL of

weqwewe [10]

Answer:

25.76 L

Explanation:

Given, Volume of Copper = 3.56 cm³

Density = 8.95 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 8.95 g/cm³ * 3.56 cm³ = 31.862 g

Mass of copper = 31.862 g

Molar mass of copper = 63.546 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{31.862\ g}{63.546\ g/mol}$

Moles of copper = 0.5014 moles

Given, Volume of nitric acid solution = 200 mL = 200 cm³

Density = 1.42 g/cm³

Considering the expression for density as:

$Density=\frac {Mass}{Volume}$

So,

So, Mass= Density * Volume = 1.42 g/cm³ * 200 cm³ = 284 g

Also, Nitric acid is 68.0 % by mass. So,

Mass of nitric acid = $\frac {68}{100}\times 284\ g$ = 193.12 g

Molar mass of nitric acid = 63.01 g/mol

The formula for the calculation of moles is shown below:

$moles = \frac{Mass\ taken}{Molar\ mass}$

Thus,

$Moles= \frac{193.12\ g}{63.01\ g/mol}$

Moles of nitric acid = 3.0649 moles

According to the reaction,

$Cu_{(s)}+4HNO_3_{(aq)}\rightarrow Cu(NO_3)_2_{(aq)} + 2NO_2_{(g)} + 2H_2O_{(l)}$

1 mole of copper react with 4 moles of nitric acid

Thus,

0.5014 moles of copper react with 4*0.5014 moles of nitric acid

Moles of nitric acid required = 2.0056 moles

Available moles of nitric acid = 3.0649 moles

Limiting reagent is the one which is present in small amount. Thus, nitric acid is present in large amount, copper is the limiting reagent.

The formation of the product is governed by the limiting reagent. So,

1 mole of copper on reaction forms 2 moles of nitrogen dioxide

So,

0.5014 mole of copper on reaction forms 2*0.5014 moles of nitrogen dioxide

Moles of nitrogen dioxide = 1.0028 moles

Given:

Pressure = 735 torr

The conversion of P(torr) to P(atm) is shown below:

$P(torr)=\frac {1}{760}\times P(atm)$

So,

Pressure = 735 / 760 atm = 0.9632 atm

Temperature = 28.2 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (28.2 + 273.15) K = 301.35 K

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

Applying the equation as:

0.9632 atm × V = 1.0028 mol × 0.0821 L.atm/K.mol × 301.35 K

⇒V = 25.76 L

4 0

3 years ago

At a certain temperature, Kc = 0.0500 and ∆H = +39.0 kJ for the reaction below, 2 MgCl2(s) + O2(g) → 2MgO(s) + Cl2(g) Calculate

Minchanka [31]

Explanation:

Since, it is shown that the reaction has been reversed. Therefore, value of $K_{c}$ will become $\frac{1}{K_{c}}$ .

Hence, new $K_{c'} = \frac{1}{K_{c}}$

= $\frac{1}{0.0500}$

= 20

Also, the number of moles of each reactant has been halved. So, $K_{c''}$ for the reaction $MgO(s) + \frac{1}{2}Cl2(g) → MgCl_{2}(s) + \frac{1}{2} O2(g)$ will also get halved.

Therefore, $K_{c''}$ = $K_{c'}$ = $(20)^{0.5}$

= 4.47

As the value of $\Delta H$ is given as +39.0 kJ. So, it means that the reaction is endothermic in nature. So, energy of reactants will be more than the products. Hence, according to Le Chatelier's principle reaction will move in the forward direction.