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3 years ago

In carbon dioxide (CO2), how many grams of oxygen (O) would there be if there are 24 grams of carbon (C)?

Chemistry

1 answer:

Zarrin [17]3 years ago

6 0

Answer:

64g of Oxygen

Explanation:

Firstly, carbon and oxygen react to form CO2 as follows;

C + O2 => CO2

According to the equation, one mole of C reacts with one mole of O2 to give one mole of CO2

Since molar mass of C= 12g/mol, O= 16g/mol

One mole of (C) contains 12g

One mole of O2 contains 16(2) = 32g

Hence, If there are 24g of C, there would be 24 × 32 / 12

= 768/12

= 64g

Hence, there would be 64g of oxygen if there are 24g of Carbon

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Read 2 more answers

Consider this reaction at equilibrium at a total pressure P1: 2SO2(g) + O2(g) â†’ 2SO3(g) Suppose the volume of this system is c

oksian1 [2.3K]

Answer:

The new equilibrium total pressure will be increased to one-half to initial total pressure.

Explanation:

From the information given :

The equation of the reaction can be represented as;

$2SO_{2(g)}+O_{2(g)} \to2SO_{3(g)}$

From above equation:

2 moles of sulphur dioxide reacts with 1 mole of oxygen (i.e 2 moles +1 mole =3 moles ) to give 2 moles of sulphur trioxide

So; suppose the volume of this system is compressed to one-half its initial volume and then equilibrium is reestablished.

So if this process takes place ; the equilibrium will definitely shift to the side with fewer moles , thus the equilibrium will shift to the right. As such; there is increase in pressure.

Let the total pressure at the initial equilibrium be $P_1$

and the total pressure at the final equilibrium be $P_2$

According to Boyle's Law; Boyle's Law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Thus;

P ∝ 1/V

P = K/V

PV = K

where K = constant

So;

PV = constant

Hence;

$P_1V_1 = P_2V_2$