How many O2 are there in 4.4 gm of Co2?

Answer 1

== == == == Standard temperature and pressure (STP), as defined by IUPAC, is 0 degrees Celsius (273 Kelvin) and 1 atmosphere of pressure.

Use the ideal gas law:

P V = n R T

where P is pressure in atmospheres, V is volume in liters, n is moles of gas, R is the universal gas constant, and T is temperature in Kelvin.

(1) (44.8) = n (0.0821) (273)

Substitute values in, and solve for moles.

n = approximately 1.9988132 moles.

With this new information, we can get our answer.

Multiply moles by oxygen's molar mass. In our case, this value is 32, as oxygen gas exists in diatomic molecules (meaning there's two oxygen atoms per molecule).

1.9988132 * 32 = 63.9620224 grams of oxygen.

Here is your answer. Apply significant figures as necessary. Look at the related question, "How do you solve Ideal Gas Law problems?" for more information. Credit for that answer goes to JEK, as of this writing.

Answer 2

I forget the easier way, so we do it this way. Use PV = nRT and find moles O2.

(1 atm)(44.8 liters) = n(0.08206 L*atm/mol *K)(298.15 K)

= 1.831 moles O2

1.831 moles O2 (32.0 grams/1 mole O2)

= 58.6 grams of O2

Answer 3

That's certainly going to depend on what substances and how many of each

are present inside the container before anything forms, and on what kind of

chemical processes take place in there.

Answer 4

In 44 g of CO2, there are 2 moles of O2 (since 1 mole of CO2 contains 2 moles of O atoms).

To find the number of moles in 4.4 g of CO2: (4.4 g CO2) / (44 g/mol CO2) = 0.1 mol CO2

Therefore, in 4.4 g of CO2, there are 0.2 moles of O2.

Answer 5

4.4 gm of CO2 is 0.1 mole.

1 mole CO2 contains 1 mole O2 molecules.

Therefore, 0.1 mole contains 0.1 mole of O2 molecules i.e. 6.022 x 1022 molecules of Oxygen

Answer 6

134,4 g oxygen are needed.

Answer 7

2.4g

Answer 8

5.6 moles