What volume of CO2 (g) measured at STP, can be obtained by the reaction of 68.0 g CaCo3 with excess hydrochloric acid? CaCO3 (s) + 2HCl (aq)    ->      CaCl2 (aq) + CO2 (g) + H2O

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`CaCO_3+2HCl rarr CaCl_2+CO_2+H_2O`

Mole ratio:

`CaCO_3:CO_2 = 1:1`

Molar mass of `CaCO_3 = 100g/(mol)`

Amount of `CaCO_3` reacted `= 68/100 = 0.68mol`

Therefore

Amount of `CO_2` formed `= 0.68mol`

At STP volume of 1mol of gas substance is `22.4l` .

Therefore volume of 0.68mol `CO_2 =...

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`CaCO_3+2HCl rarr CaCl_2+CO_2+H_2O`

 

Mole ratio:

`CaCO_3:CO_2 = 1:1`

 

Molar mass of `CaCO_3 = 100g/(mol)`

Amount of `CaCO_3` reacted `= 68/100 = 0.68mol`

 

Therefore

Amount of `CO_2` formed `= 0.68mol`

 

At STP volume of 1mol of gas substance is `22.4l` .

Therefore volume of 0.68mol `CO_2 = 22.4xx0.68 = 15.232l`

 

So the answer is `15.232l`

 

Assumption

`CO_2` act as an ideal gas.

Approved by eNotes Editorial Team