Question

What volume of carbon dioxide is produced when 0.489 mol of calcium carbonate reacts completely according to the following reaction at 0°C and 1 atm? calcium carbonate ( s ) calcium oxide ( s ) + carbon dioxide ( g )

Answer 1

11.0L of carbon dioxide is produced

Step-by-step explanation:

Balanced equation:
`CaCO_(3)(s)\rightarrow CaO(s)+CO_(2)(g)`

According to balanced equation, 1 mol of
`CaCO_(3)` produces 1 mol of
`CO_(2)`

So, 0.489 mol of
`CaCO_(3)` produces 0.489 mol of
`CO_(2)`

Let's assume
`CO_(2)` behaves ideally.

So,
`P_{CO_(2)}V_{CO_(2)}=n_{CO_(2)}RT`

where P is pressure, V is volume , n is number of moles, R is gas constant and T is temperature in kelvin

Plug-in all the values in the above equation-

`(1atm)* V_{CO_(2)}=(0.489mol)* (0.0821L.atm.mol^(-1).K^(-1))* (273K)`

or,
`V_{CO_(2)}=11.0L`

So, 11.0L of carbon dioxide is produced