Question

Calculate the volume of gas that can be obtained from reacting 1.8 g of calcium carbonate with dilute nitric acid at R.T.P? (1 mole of gas occupies 24 000 cm3 at R.T.P )

Answer 1

`CaCO_3(s) + 2HNO_3(aq) rarr Ca(NO_3)_2(aq)+CO_2(g)uarr`

Explanation:

And thus there is 1:1 equivalence between the moles of calcium carbonate, and the moles of carbon dioxide gas evolved. As always, writing the stoichiometric equation is the necessary preliminary.....

`"Moles of calcium carbonate"=(1.80*g)/(100.09*g*mol^-1)=0.018*mol`..

And so we get ............................

`0.018*molxx24xx10^3*cm^3*mol^-1=432*cm^3` with respect to evolved `CO_2`...