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4 years ago

What pressure is required to achieve a co2 concentration of 7.90×10−2 m at 20∘c?

Chemistry

1 answer:

Savatey [412]4 years ago

7 0

Answer:- 1.90 atm

Solution:- It is based on combined gas law equation, PV = nRT

In this equation, P is pressure, V is volume, n is moles of gas, R is universal gas constant and T is kelvin temperature.

If we divide both sides by V then:

$P=\frac{nRT}{V}$

We know that, molarity is moles per liter. So, in the above equation we could replace $\frac{n}{V}$ by molarity, M of the gas. The equation becomes:

P = MRT

T = 20 + 273 = 293 K

M = $7.9*10^-^2$

Let's plug in the values in the equation:

P = $(7.9*10^-^2)(0.0821)(293)$

P = 1.90 atm

So, the pressure of the gas is 1.90 atm.

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For each of the reactions, calculate the mass (in grams) of the product formed when 15.12g of the underlined reactant completely

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Answer:

Part A : amount of product (KCl) = 28.88 g

Part B : amount of product (KBr) = 46.13 g

Part C : amount of product (Cr₂O₃) = 17.3 g

Part D: amount of product (SrO) = 35.76 g

Explanation:

Part A:

Data Given:

Reaction :

2K(s) + Cl₂(g) --------> 2KCl

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is chlorine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Cl₂(g) --------> 2KCl

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KCl

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KCl = (39 + 35.5)

Molar mass of KCl = 74.5 g/mol

So the look again to reaction in terms of grams

2K(s) + Cl₂(g) --------> 2KCl

2mole (39 g/mol) 2mole (74.5 g/mol)

78 g 149 g

Apply the Unity formula

78 g of Potassium ≅ 149 g of KCl

Then

15.12 g of Potassium ≅ how many g of Product (KCl)

By doing cross multiplication

X g of Product (KCl) = 149 g of KCl x 15.12 g of K / 78 g of K

X g of Product (KCl) = 28.88 g

So the amount of product (KCl) = 28.88 g

_________________________________________

Part B:

Data Given:

Reaction :

2K(s) + Br₂(g) --------> 2KBr

Amount of underline Reactant (K) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Potassium (K) is 15.12g and other reactant that is Bromine is more than enough so the K is limiting reagent.

So, amount of product depend on the amount of Potassium (K)

Now Look at Given Reaction:

2K(s) + Br₂(g) --------> 2KBr

2mol 1mol 2mol

it shows that

2 mole of K give 2 mole of KBr

if we represent mole in grams

Then

Molar mass of K = 39 g/mol

Molar mass of KBr = (39 + 80)

Molar mass of KBr = 119 g/mol

So, look again to reaction in terms of grams

2K(s) + Br₂(g) --------> 2KBr

2mole (39 g/mol) 2mole (119 g/mol)

78 g 238 g

Apply the Unity formula

78 g of Potassium ≅ 238 g of KBr

Then

15.12 g of Potassium ≅ how many g of Product (KBr)

By doing cross multiplication

X g of Product (KBr) = 238 g of KBr x 15.12 g of K / 78 g of K

X g of Product (KBr) = 46.13 g

So the amount of product (KBr) = 46.13 g

__________________________________________

Part C:

Data Given:

Reaction :

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

Amount of underline Reactant (Cr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Chromium (Cr) is 15.12g and other reactant that is Oxygen is more than enough so the Cr is limiting reagent.

So, amount of product depend on the amount of Chromium (Cr)

Now Look at Given Reaction:

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4mol 3mol 2mol

it shows that

4 mole of Cr give 2 mole of Cr₂O₃

if we represent mole in grams

Then

Molar mass of Cr = 52 g/mol

Molar mass of 2Cr₂O₃ = 2 [2 (52) + 3(16) ] = 2 (104+ 48)

Molar mass of 2Cr₂O₃ = 304 g/mol

So, look again to reaction in terms of grams

4Cr(s) + 3O₂(g) --------> 2Cr₂O₃

4 mol (52 g/mol) 2 mole (304 g/mol)

208 g 608 g

Apply the Unity formula

208 g of Chromium ≅ 608 g of Cr₂O₃

Then

15.12 g of Chromium ≅ how many g of Product (Cr₂O₃)

By doing cross multiplication

X g of Product (Cr₂O₃) = 238 g of Cr₂O₃ x 15.12 g of Cr / 208 g of Cr

X g of Product (Cr₂O₃) = 17.3 g

So the amount of product (Cr₂O₃) = 17.3 g

________________________________________

Part D:

Data Given:

Reaction :

2Sr(s) + O₂(g) --------> 2SrO(s)

Amount of underline Reactant (Sr) = 15. 12g

amount of other reactant = more than enough

Explanation:

As the Strontium (Sr) is 15.12g and other reactant that is Oxygen is more than enough so the Sr is limiting reagent.

So, amount of product depend on the amount of Strontium (Sr)

Now Look at Given Reaction:

2Sr(s) + O₂(g) --------> 2SrO(s)

2mol 1mol 2mol

it shows that

2 mole of Sr give 2 mole of SrO

if we represent mole in grams

Then

Molar mass of Sr = 87.6 g/mol

Molar mass of 2SrO = 2 [87.6 + 16] = 2 (103.6)

Molar mass of 2SrO = 207.2 g/mol

So, look again to reaction in terms of grams

2Sr(s) + O₂(g) --------> 2SrO(s)

2 mol ( 87.6 g/mol) 2 mole (207.2 g/mol)

175.2 g 414.4 g

Apply the Unity formula

175.2 g of Strontium ≅ 414.4 g of SrO

Then

15.12 g of Strontium ≅ how many g of Product (SrO)

By doing cross multiplication

X g of Product (SrO) = 414.4 g of SrO x 15.12 g of Sr / 175.2 g of Sr

X g of Product (SrO) = 35.76 g

So the amount of product (SrO) = 35.76 g

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