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2 years ago

How many moles of C3H8 will be produced from 75 moles of CO2?C3H8 + 5 O2 → 3 CO2 + 4 H2O

Chemistry

1 answer:

levacccp [35]2 years ago

8 0

25 moles of C3H8 will be produced from 75 moles of CO2.

<h3>Mole calculation</h3>

To find the value of moles of a product from the number of moles of a reactant, it is necessary to observe the stoichiometric ratio between them:

$C_3H_8 + 5 O_2 = > 3 CO_2 + 4 H_2O$

Analyzing the reaction, it is possible to see that the stoichiometric ratio is 1:3, so we can perform the following expression:

C3H8 CO2

$\frac{1mol}{xmol} =\frac{3mol}{75mol}$

$x = 25mol$

So, 25 moles of C3H8 will be produced from 75 moles of CO2.

Learn, more about mole calculation in: brainly.com/question/2845237

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3 years ago

Why a neutral atom has the same number of protons and electrons

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Explanation:

When an atom has an equal number of electrons and protons, it has an equal number of negative electric charges (the electrons) and positive electric charges (the protons). The total electric charge of the atom is therefore zero and the atom is said to be neutral. ... Chemically, we say that the atoms have formed bonds.

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4 years ago

For the decomposition of A to B and C, A(s)⇌B(g)+C(g) how will the reaction respond to each of the following changes at equilibr

lys-0071 [83]

Answer:

a. No change.

b. The equilibrium will shift to the right.

c. No change

d. No change

e. The equilibrium will shift to the left

f. The equilibrium will shift to the right

Explanation:

We are going to solve this question by making use of Le Chatelier´s principle which states that any change in a system at equilibrium will react in such a way as to attain qeuilibrium again by changing the equilibrium concentrations attaining Keq again.

The equilibrium constant for A(s)⇌B(g)+C(g)

Keq = Kp = pB x pC

where K is the equilibrium constant ( Kp in this case ) and pB and pC are the partial pressures of the gases. ( Note A is not in the expression since it is a solid )

We also use Q which has the same form as Kp but denotes the system is not at equilibrium:

Q = p´B x p´C where pB´ and pC´ are the pressures not at equilibrium.

a. double the concentrations of Q which has the same form as Kp but : products and then double the container volume

Effectively we have not change the equilibrium pressures since we know pressure is inversely proportional to volume.

Initially the system will decrease the partial pressures of B and C by a half:

Q = pB´x pC´ ( where pB´and pC´are the changed pressures )

Q = (2 pB ) x (2 pC) = 4 (pB x PC) = 4 Kp ⇒ Kp = Q/4

But then when we double the volume ,the sistem will react to double the pressures of A and B. Therefore there is no change.

b. double the container volume

From part a we know the system will double the pressures of B and C by shifting to the right ( product ) side since the change reduced the pressures by a half :

Q = pB´x pC´ = ( 1/2 pB ) x ( 1/2 pC ) = 1/4 pB x pC = 1/4 Kp

c. add more A

There is no change in the partial pressures of B and C since the solid A does not influence the value of kp

d. doubling the concentration of B and halve the concentration of C

Doubling the concentrantion doubles the pressure which we can deduce from pV = n RT = c RT ( c= n/V ), and likewise halving the concentration halves the pressure. Thus, since we are doubling the concentration of B and halving that of C, there is no net change in the new equilibrium:

Q = pB´x pC´ = ( 2 pB ) x ( 1/2 pC ) = K

e. double the concentrations of both products

We learned that doubling the concentration doubles the pressure so:

Q = pB´x pC´ = ( 2 pB ) x ( 2 pC ) = 4 Kp

Therefore, the system wil reduce by a half the pressures of B and C by producing more solid A to reach equilibrium again shifting it to the left.

f. double the concentrations of both products and then quadruple the container volume

We saw from part e that doubling the concentration doubles the pressures, but here afterward we are going to quadruple the container volume thus reducing the pressure by a fourth:

Q = pB´x pC´ = ( 2 pB/ 4 ) x (2 pC / 4) = 4/16 Kp = 1/4 Kp

So the system will increase the partial pressures of B and C by a factor of four, that is it will double the partial pressures of B and C shifting the equilibrium to the right.

If you do not see it think that double the concentration and then quadrupling the volume is the same net effect as halving the volume.

3 0

3 years ago

Question 6: The Ideal Gas Law (7 points) a. What is the mathematical equation for the ideal gas law? Identify each variable and

azamat

The mathematical equation for the ideal gas law is PV = nRT.

<h3>What is an ideal gas equation?</h3>

The ideal gas law (PV = nRT) relates the macroscopic properties of ideal gases. An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).

(A)

PV = nRT

The ideal gas equation is formulated as: PV = nRT. In this equation, P refers to the pressure of the ideal gas, V is the volume of the ideal gas, n is the total amount of ideal gas that is measured in terms of moles, R is the universal gas constant, and T is the temperature.

(B)

Where the pressure - P, is in atmospheres (atm) the volume - V, is in liters (L) the moles -n, are in moles (m) and Temperature -T is in Kelvin (K) as in all gas law calculations.

(C)

A. Boyle's law

B. Charles's law

C. Avogadro's law

D. Dalton's law

____ $- P_1V_1$ = $P_2 V_2$