URGENT! I need someone to check my calculation...

Given 8KClO3 + C12H22O11 --> 8KCl + 12CO2 + 11H2O ( I balanced the equation already)

(a) How many grams of KClO3 are needed to react with 6.15 x 10^22 molecules of C12H22O11?

n of C12H22O11,
=(6.15 X 10^22)/ (6.02 X 10^23)
= 0.102 mol

n of KClO3,
= 8 x 0.1022
= 0.8176 mol

Mr of KClO3,
= (39.1) + 35.5 +(16 x 0.3)
=122.6 g/mol

Mass of KClO3,
= 0.8176 X 122.6
=100.2g

(c) Calculate the percentage yield of the reaction if 4.6g of KClO3 gives 4.7 g of CO2

n KClO3,
=4.6/122.6
=0.0375
=0.0375/8
=4.6875 X 10^-3

Mass of KClO3,
=4.6875 X 10^-3 x 122.6
=0.5746 g

n CO2,
=4.7/ 44
=0.1068
=0.1068/12
=8.9 X 10^-3

mass CO2,
=44 X ( 8.9 X 10^-3)
=0.3916 g

thus percentage yield,
=0.5746 - 0.3916
=0.183 g
=0.183/0.3916 x (100)
=46.7%

(c) Calculate the percentage yield of the reaction if 4.6g of KClO3 gives 4.7 g of CO2

KClO3 to CO2 is 8 to 12 (or 2 to 3)

2 is to 3 as 0.0375 mol is to x

x = 0.05625 mol of CO2 produced

0.05625 mol times 44 g/mol = 2.475 g of CO2 produced

You can do the percent yield.

You have a misapprehension about how to do stoichiometry. You divided the KClO3 by 8 and the CO2 by 12. That creates a 1:1 ratio of amounts that means nothing. This is because, for every 2 units of KClO3 that react, 3 units of CO2 are produced. In other words, 0.05625 is 1.5 times greater than 0.0375.

Look here:

http://www.chemteam.info/Stoichiometry/M…

Perhaps do some searching for videos on this topic. Look for something from Kahn Academy. He has lots of good stuff. Try kentchemistry as well.