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?

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Answer 1 · 2015-02-23T13:24:19

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#