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?

1 Answer
Feb 23, 2015

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=40A_rC=12 A_rO=16

and I mole of gas occupies 22.4l @ stp:

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

100grarr22.4l

So 1l requires 100/22.4g

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