Question

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

Answer 1

Answer:

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

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 ×$\frac{1 mol}{100,09 g}$ =

9,53 CaCO₃ moles ≡ CO₂ moles

9,53 CO₂ moles ×$\frac{44,01 g}{1 mol}$ = 419,4 g of CO₂

Thus, density of Carbon dioxide is:

$\frac{419,4 g}{225 L}$ = 1,86 g/L

I hope it helps!