A volume of #5.6*dm^3# #CO_2(g)# is evolved from WHAT mass of solid calcium carbonate under standard conditions?

1 Answer
anor277
Jul 27, 2017

Approx. #20*g# of stuff.......

Explanation:

We follow the stoichiometric equation.....

#CaCO_3(s) +Delta rarr CaO(s) +CO_2(g)#.

We know that under standard conditions of temperature, and pressure, #1*mol# of Ideal Gas occupies a volume of #25.4*dm^3#; you will have to fill in the appropriate standard according to your syllabus.

And thus mass equivalence of SOLID CALCIUM CARBONATE.....

#=(5.6*dm^3)/(25.4*dm^3*mol^-1)xx100.09*g*mol^-1=??*g#

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