Determine the volume of o2 (at stp) formed when 50.0 g of kclo3 decomposes according to the following reaction. the molar mass for kclo3 is 122.55 g/mol. 22kclo3 (s) --> 2 kcl(s) + 3 o2(g)

Taking into account the reaction stoichiometry, ideal gas law and the definition of STP conditions, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

The balanced reaction is:

2 KClO₃ → 2 KCl + 3 O₂

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

The molar mass for KClO₃ is 122.55 g/mol. Then, by reaction stoichiometry, the following mass quantity of the compound participates in the reaction:

245.1 grams of KClO₃

Then it is possible to apply the following rule of three: if by reaction stoichiometry 245.1 grams of KClO₃ form 3 moles of O₂, 50 grams of KClO₃ form how many moles of O₂?

amount of moles of O₂= 0.612 moles

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas.

On the other side, the STP conditions refer to the standard temperature and pressure. Pressure values at 1 atmosphere and temperature at 0 ° C are used and are reference values for gases.

Then, in this case you know:

Replacing in the ideal gas:

1 atm× V= 0.612 moles× 0.082 × 273 K

Solving:

V=13.70 L

Finally, 13.70 L of O₂ (at STP) is formed when 50.0 g of KClO₃ decomposes according to the following reaction.

Learn more:

STP conditions refer to the standard temperature and pressure. The volume of the oxygen at STP will be 13.70 L when 50 grams of potassium chlorate decomposes.

The chemical equation of decomposition if potassium chlorate can be written as:

From the stoichiometry of the reaction, it can be observed as:

Now, calculating the mass of potassium chlorate participating in the reaction:

Now, substituting the values to calculate the moles of oxygen:

From the ideal gas equation:

P is pressure, Tis  the temperature, and the volume, V. R is ideal gas constant.

Substituting the values:

Therefore, the 13.70 L of oxygen at STP will be formed when 50 grams of potassium chlorate decomposes into potassium and oxygen.

To know more about ideal gas equation, refer to the following link:

link

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.