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

Answer:

11.0L of carbon dioxide is produced

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-

$(1atm)\times V_{CO_{2}}=(0.489mol)\times (0.0821L.atm.mol^{-1}.K^{-1})\times (273K)$

or, $V_{CO_{2}}=11.0L$

So, 11.0L of carbon dioxide is produced