posted by Pat on .
A mixture containing KCLO3, K2CO3, KHCO3, and KCL was heated, producing CO2, O2, and H2O (water) gases according following equations:
2KCLO3 -> 2KCL + O2
2KHCO3 -> K2O + H2O + 2CO2
2CO3 -> K2O + CO2
The KCL does not reacnder the conditions of the reaction. If 100.0g of the mixture produces 1.8g of H20, 13.2g of CO2, and 4g of O2, what was the compttion of the original mixture?(assume complete decomposition of the mixture)
someone please show me how to do this!
* all O'in the equations stand for oxygen, nothing means 20, its 2O, 2 then oxygen comes after.
O2 came from equation 1, so change the grams O2 to moles, then compute the moles and mass of potassium chlorate.
Water came only from equation 2, change the water mass to moles, and compute the moles/mass of potassium hydrogencarbonate. Also, compute the moles/mass of CO2 given off, subtract that from 13.2g. The remainder is given off in reaction 3. Repeat mole/mass relations to get the amount of K2CO3.
For the original amount of KCL, add the masses you found for KCLO3, KHCO3, and K2CO3, then subtract from 100g.
what is the name of the following compound K2O