Question

A 435.0 g sample of CaCO3 that is 95% pure gives 225L of CO2 when reacted with excess of hydrochloric acid. What is the density in g/L of the CO2 given the reaction equation below? CaCO3(s)+2HCl(I)-CaCl2(s)+CO2(g)+H2O(I)

Answer 1

Density = 0.82 gm/L


Explanation:
Caco3= 40+12+16x2
= 100gm

CO2= 12+16x2
=44 gm

44/100 x 435

= 191.4gm


The mass of CO2 = 191.4gm
Volume of co2= 225L


Density= mass/volume = 191.4gm/225 L


= 0.85 gm/L

Answer 2

Density = 0.85 gm/L

Step-by-step explanation:

Check the attached image for solution.