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3 years ago

How many moles are in 297 g of NH3?

Chemistry

1 answer:

BigorU [14]3 years ago

8 0

Answer:

1. 17.4 moles.

2. 1.13 moles

3. 315.5 moles

4. 390.6g

5. 1.13x10⁶ moles

6. 14.8 moles

7. 337 moles

8. 2.15x10²⁴ molecules

9. 3.13x10²⁴ atoms

10. 1.38x10²⁴ particles

11. 1517g

12. 455g of CaF₂

Explanation:

We can convert formula units to moles or vice versa using Avogadro's number and moles to grams using molar mass of the substance:

1. Molar mass NH3: 17.031g/mol

297g * (1mol / 17.031g) = 17.4 moles

2. Molar mass MgCO3: 84.3g/mol

95g * (1mol / 84.3g) = 1.13 moles

3. Using Avogadro's number (6.022x10²³ formula units / mol):

1.9x10²⁶FU * (1mol / 6.022x10²³FU) = 315.5 moles

4. Molar mass H2O: 18g/mol

21.7mol * (18g / mol) = 390.6g

5. Using Avogadro's number (6.022x10²³ molecules / mol):

6.78x10²⁹molecules * (1mol / 6.022x10²³FU) = 1.13x10⁶ moles

6. 8.9x10²⁴FU * (1mol / 6.022x10²³FU) = 14.8 moles

7. Using Avogadro's number (6.022x10²³ atoms / mol):

2.03x10²⁶atoms* (1mol / 6.022x10²³FU) = 337 moles

8. 3.569mol * (6.022x10²³ molecules / 1mol) = 2.15x10²⁴ molecules

9. 5.2mol * (6.022x10²³ atoms / 1mol) = 3.13x10²⁴ atoms

10. Molar mass Li₂SO₄: 109.94g/mol:

36g * (1mol / 109.94g) * (6.022x10²³ molecules / 1mol) * (7 particles / 1molecule) = 1.38x10²⁴ particles

<em>Assuming particles are atoms and in 1 molecule of Li₂SO₄ you have 7 atoms.</em>

11. Molar mass Cl₂: 70.9g/mol:

21.4mol * (70.9g / mol) = 1517g

12. Molar mass CaF₂: 78.07g/mol:

3.51x10²⁴FU * (1mol / 6.022x10²³FU) * (78.07g / mol) = 455g of CaF₂

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calculate the mass of calcium phosphate and the mass of sodium chloride that could be formed when a solution containing 12.00g o

Leviafan [203]

Answer : The mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

Explanation : Given,

Mass of $Na_3PO_4$ = 12.00 g

Mass of $CaCl_2$ = 10.0 g

Molar mass of $Na_3PO_4$ = 164 g/mol

Molar mass of $CaCl_2$ = 111 g/mol

Molar mass of $NaCl$ = 58.5 g/mol

Molar mass of $Ca_3(PO_4)_2$ = 310 g/mol

First we have to calculate the moles of $Na_3PO_4$ and $CaCl_2$ .

$\text{Moles of }Na_3PO_4=\frac{\text{Given mass }Na_3PO_4}{\text{Molar mass }Na_3PO_4}$

$\text{Moles of }Na_3PO_4=\frac{12.00g}{164g/mol}=0.0732mol$

and,

$\text{Moles of }CaCl_2=\frac{\text{Given mass }CaCl_2}{\text{Molar mass }CaCl_2}$

$\text{Moles of }CaCl_2=\frac{10.0g}{111g/mol}=0.0901mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is:

$2Na_3PO_4+3CaCl_2\rightarrow 6NaCl+Ca_3(PO_4)_2$

From the balanced reaction we conclude that

As, 3 mole of $CaCl_2$ react with 2 mole of $Na_3PO_4$

So, 0.0901 moles of $CaCl_2$ react with $\frac{2}{3}\times 0.0901=0.0601$ moles of $Na_3PO_4$

From this we conclude that, $Na_3PO_4$ is an excess reagent because the given moles are greater than the required moles and $CaCl_2$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaCl$ and $Ca_3(PO_4)_2$

From the reaction, we conclude that

As, 3 mole of $CaCl_2$ react to give 6 mole of $NaCl$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{6}{3}\times 0.0901=0.1802$ mole of $NaCl$

and,

As, 3 mole of $CaCl_2$ react to give 1 mole of $Ca_3(PO_4)_2$

So, 0.0901 mole of $CaCl_2$ react to give $\frac{1}{3}\times 0.0901=0.030$ mole of $Ca_3(PO_4)_2$

Now we have to calculate the mass of $NaCl$ and $Ca_3(PO_4)_2$

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

$\text{ Mass of }NaCl=(0.1802moles)\times (58.5g/mole)=10.5g$

and,

$\text{ Mass of }Ca_3(PO_4)_2=\text{ Moles of }Ca_3(PO_4)_2\times \text{ Molar mass of }Ca_3(PO_4)_2$

$\text{ Mass of }Ca_3(PO_4)_2=(0.030moles)\times (310g/mole)=9.3g$

Therefore, the mass of calcium phosphate and the mass of sodium chloride that formed could be, 9.3 and 10.5 grams respectively.

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99% of matter that makes up living organisms consists of how many elements?

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Explanation:

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4 years ago

The question in the picture, I really a correct answer, no cheap answers

Delvig [45]

Answer:

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Explanation:

We'll begin by converting 350 mL to L. This can be obtained as follow:

1000 mL = 1 L

Therefore,

350 mL = 350 mL × 1 L /1000 mL

350 mL = 0.35 L

Next, we shall determine the number of mole of KC₂H₃O₂ in the solution. This can be obtained as follow:

Volume = 0.35 L

Molarity of KC₂H₃O₂ = 2.75 M

Mole of KC₂H₃O₂ =?

Molarity = mole /Volume

2.75 = Mole of KC₂H₃O₂ / 0.35

Cross multiply

Mole of KC₂H₃O₂ = 2.75 × 0.35

Mole of KC₂H₃O₂ = 0.9625 mole

Finally, we shall determine the mass of KC₂H₃O₂ needed to prepare the solution. This can be obtained as illustrated below:

Mole of KC₂H₃O₂ = 0.9625 mole

Molar mass of KC₂H₃O₂ = 39 + (12×2) +(3×1) + (16×2)

= 39 + 24 + 3 + 32

= 98 g/mol

Mass of KC₂H₃O₂ =?

Mass = mole × molar mass

Mass of KC₂H₃O₂ = 0.9625 × 98

Mass of KC₂H₃O₂ = 94.325 g

Thus, the mass of KC₂H₃O₂ needed to prepare the solution is 94.325 g

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