CaCO3+2HCl->CO2+H2O. Experiment gave actual yield of 1.000x10^2g of CO2. Began with 2.00x10^2 of HCl and ample CaCO3 left unreacted, what was the percent of yield? This means that HCl...
CaCO3+2HCl->CO2+H2O. Experiment gave actual yield of 1.000x10^2g of CO2. Began with 2.00x10^2 of HCl and ample CaCO3 left unreacted, what was the percent of yield?
This means that HCl was the limiting reactant, I assume.
1 Answer
jeew-m | College Teacher | (Level 1) Educator Emeritus
`CaCO_3+2HCl rarr CO_2+H_2O`
Molic weight of `CO_2` = 44`g/(mol)`
Molic weight of HCl `= 36.5g/(mol)`
Actual yield of `CO_2`
`= 1xx10^2g`
`= 100g`
`= 100/44`
`= 2.27 mol`
Reacted HCl
`= 2xx10^2g`
`= 200g`
`= 200/36.5 mol`
`= 5.48 mol`
Mole ratio
`HCl:CO_2 = 2:1`
Theoretical `CO_2` that should yield `= 5.48/2 = 2.74 mol`
Mass of theoretical `CO_2` that should yield `= 2.74xx44 = 120.56g`
Percent of yield of `CO_2`
`= 100/120.56xx100%`
`= 82.95%`
So the percent yield of `CO_2` is 82.95%.
Reasons for loss in percentage of `CO_2`
- `CaCO_3` may not pure and there may have been impurities
- `CO_2` is not collected properly and some of `CO_2` may have lost in collecting process