Exactly 12.0 grams of potassium chlorate is placed in a reaction vessel; following the decomposition, 1.55g of kclo3- remain unreacted. calculate the total volume of oxygen collected at 25 degrees celsius and 0.95atm.

3.3L

Explanation:

Firstly, we write the reaction equation :

2KClO3 ——-> 2KCl + 3O2

Now we need to know the mass of potassium chlorate reacted = total mass - mass of unreacted chlorate

= 12.0 - 1.55 = 10.45g

Now we need to know the number of moles of chlorate reacted. This equals the mass reacted divided by the molar mass.

The molar mass of potassium chlorate = 39 + 35.5 + 3(16) = 122.5g/mol

The number of moles = 10.45/122.5= 0.0853moles.

From the reaction equation, we know that 2 moles of chlorate yielded 3 moles of oxygen. The number of moles of oxygen produced from 0.0853moles chlorate is thus = 0.12795moles

Now, to calculate the total volume of oxygen collected, we use the ideal gas equation.

Here PV = nRT

hence, V = nRT/P

We have the following values: P = 0.95atm

If 1 atm = 101,325Pa, 0.95atm = 101325 * 0.95 = 96258.75Pa

T = 25 degrees Celsius = 25 + 273.15 = 298.15K

n = 0.12795 moles

R = molar gas constant = 8314.462L.Pa/K.mol

Putting these values into the equation: V = (0.12795 * 8314.462 * 298.15)/ 96,258.75 = 3.3L

the answers would be

the first one and the second one