# 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.

## Expert Answers

`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
Approved by eNotes Editorial Team

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