Question

I have the answer key for this but I need a walkthrough. The balanced equation for calcium carbonate into lime is: CaCO3 + heat ----> CaO + CO2 How many grams of calcium carbonate must be decomposed to produce 5.0 L of carbon dioxide gas at STP?

Answer 1

22.32g are required.

`CaCO_(3(s))rarrCaO_((s))+CO_(2(g))`

So 1 mole `CaCO_3` gives 1 mole `CO_2`

Convert to grams;

`A_rCa=40` `A_rC=12` `A_rO=16`

and I mole of gas occupies `22.4l` @ stp:

`[40+12+(3xx16)rarr22.4l`

`100grarr22.4l`

So `1l` requires `100/22.4`g

So `5l` requires `(100)/(22.4)xx5=22.32g`