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.

$C a C O_{3 (s)} \to C a O_{(s)} + C O_{2 (g)}$ $CaCO_(3(s))rarrCaO_((s))+CO_(2(g))$

So 1 mole $C a C O_{3}$ $CaCO_3$ gives 1 mole $C O_{2}$ $CO_2$

Convert to grams;

$A_{r} C a = 40$ $A_rCa=40$ $A_{r} C = 12$ $A_rC=12$ $A_{r} O = 16$ $A_rO=16$

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

$[40 + 12 + (3 \times 16) \to 22.4 l$ $[40+12+(3xx16)rarr22.4l$

$100 g \to 22.4 l$ $100grarr22.4l$

So $1 l$ $1l$ requires $\frac{100}{22.4}$ $100/22.4$ g

So $5 l$ $5l$ requires $\frac{100}{22.4} \times 5 = 22.32 g$ $(100)/(22.4)xx5=22.32g$