Under which conditions are gases most soluble in water?

When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Temka [501]

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

4 0

3 years ago

A photon has an energy of 1.10 x 10 -13 j. What is the photon's wavelength? What type of electromagnetic radiation is it?

lawyer [7]

Answer:

λ = 0.002 nm

The given photon is either x-ray or gamma ray because the range of x-ray and gamma ray is 1 nm-0.1 pm.

Explanation:

Given data:

Energy of photon = 1.10 × 10⁻¹³ J.

Wavelength of photon = ?

Solution:

Formula:

E = h.c / λ

λ = h. c / E

λ = 6.626 × 10⁻³⁴ j. s × 3×10⁸ m/s / 1.10 × 10⁻¹³ J.

λ = 19.878 × 10⁻²⁶m / 1.10 × 10⁻¹³ J

λ = 18.071 × 10⁻¹³ m

λ = 18.071 × 10⁻¹³ × 10⁹

λ = 18.071 × 10⁻⁴ nm

λ = 0.002 nm

The given photon is either x-ray or gamma ray because the range of x-ray and gamma ray is 1 nm-0.1 pm.

3 0

4 years ago

Which has the greatest mass? 1 mole Ca 1 mole CO 1 mole O2 1 mole CH4 1 mole NO

tino4ka555 [31]

Answer: 1 mole of $Ca$ has the greatest mass.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP , contains avogadro's number $(6.023\times 10^{23})$ of particles and weighs equal to the molecular mass of the substance.

1 mole of $Ca$ has a mass of 40 g.

1 mole of $CO$ has a mass of 28 g

1 mole of $O_2$ has a mass of 32 g

1 mole of $CH_4$ has a mass of 16 g.

1 mole of $NO$ has a mass of 30 g.

Thus the greatest mass is of 1 mole of $Ca$

8 0

3 years ago

Scientists often _________________________ and _______________________ to help them interpret complex ideas. *

Ray Of Light [21]

Answer:

no sir

Explanation:

8 0

3 years ago

PLEASE HELP! Class is almost ending, I don't want to repeat.

nikklg [1K]

ANSWERS:

1. $2KClO_{3}$ ⇒ 2KCl + $3O_{2}$

2. $H_{2} C_{2} O_{4} + 2NaOH$ ⇒ $Na_{2} C_{2} O_{4} + 2 H_{2} O$

3. $2 C_{10} H_{22} + 31O_{2}$ ⇒ $20C O_{2} + 22 H_{2} O$

4. $4 H_{2} + Fe_{3} O_{4}$ ⇒ $3Fe +4 H_{2} O$

5. $4P + 5 O_{2}$ ⇒ $2 P_{2} O_{5}$

I have attached my work for 1-3. I would like to see if you can get 4 and 5 on your own. but, if you are struggling/copnfused please let me know in the comments! :)

6 0

4 years ago