Please help me out with the questions above, I will be forever thankful! Unknown
ID: 488958 • Letter: P
Question
Please help me out with the questions above, I will be forever thankful!
Unknown sample (KCLO_3 + KCL) Determine the weight of unknown heated. Determine the weight of oxygen gas released using the constant weight obtained after completion of the heating's. Determine the number of oxygen gas (O_2) lost? Using stoichiometry, use the number of O_2 gas released to determine the moles of KCLO_3 decomposed. Determine the number of grams of KCLO_3 decomposed. Based on the number of grams of KCLO_3 decomposed and the total amount of unknown placed in the test tube, determine the percent KCLO_3 present within the unknown.Explanation / Answer
Solution: Preliminary steps
The given problem is based on catalytic decomposition of KClO3
On heating KClO3 in the presence of catalyst MnO2 (39.84g)
Step 1- write the balanced equation for the chemical change taking place.
2KClO32KCl+ 3O2
Step 2- Find out the molecular mass of all the compounds and element/s present in the equation
[At.mass of K=39, Cl=35.5 and O=16]
Molecular mass of
KClO3=39+35.5+(3x16)=122.5 U
KCl = 74.5 U
O2 =2x16=32 U
Molar mass is molecular mass expressed in g. Substitute these values in the equation
Step 3: Substitute these values in the equation
2KClO3 2KCl + 3O2
2moles 2 moles 3moles
2 (122.5 )g 2 (74.5)g 3 (32)g
245g 149g 96g
Calculation
Weight of the test tube is constant in all the weightings
Hence
of unknown sample)
= (39.84g- 39.72g)= 0.12g
245g of KClO3 releases = 96g of O2
Therefore 0.12g of KClO3 releases = 96g x 0.12g /245 g = of O2= 0.047 g of O2
32g of O2= 6.02x 10 23 number of O2 molecules
Therefore 0.047 g of O2 =(6.02x 10 23 x0.047g)/32g =0.00884 x 10 23 number of O2
molecules or
= 8.84 x 10 20 number of O2 molecules
number of O2 molecules lost= = 8.84 x 10 20
2moles2 moles3moles
2 (122.5 )g2 (74.5)g3 (32)g
245g149g 96g
8.84 x 10 20 number of
Molecules
6.02x 10 23 number of O2 molecules = 1 mole
Therefore 8.84 x 10 20 number of O2 molecules = (1mole x 8.84 x 10 20) /6.02x 10 23
=1.4684 x 10 3 moles of O2
3 moles of O2 is released by the decomposition of = 2 moles of KClO3
Therefore 1.4684 x 10 3 moles of O2 is released by= (2 x 1.4684 x 10 3 )/3=0.9789 x 10 3
moles of KClO3
= 9.789 x 10 4 moles of KClO3