Question

A 1004.0g sample of calcium carbonate that is 95.0% pure

gives225L of CO2 at STP when reacted with an excess
ofhydrochloric acid. What is the density (in g/L) of the
carbondioxide?

Answer 1

The density of carbon dioxide is 1,86 g/L

Step-by-step explanation:

The global reaction is:

2 HCl (aq)+ CaCO₃ (s) → CaCl₂(aq)+ H₂O(l)+ CO₂(g)

To obtain density it is necessary to obtain calcium carbonate moles -with molar mass of CaCo₃ = 100,09 g/mol- that are the same than CO₂ moles. Then, this moles must be converted to grams -CO₂ weights 44,01 g/mol- and, with the given liters (225 L) will be possible to know density, thus:

1004,0g × 95,0% = 953,8 g of CaCO₃

953,8 g of CaCO₃ ×
`(1 mol)/(100,09 g)` =

9,53 CaCO₃ moles ≡ CO₂ moles

9,53 CO₂ moles ×
`(44,01 g)/(1 mol)` = 419,4 g of CO₂

Thus, density of Carbon dioxide is:

`(419,4 g)/(225 L)` = 1,86 g/L

I hope it helps!