The balanced equation is shown below.

Question

2 HCl(aq) +CaCO3(s)=CaCl2(aq) +H2O(l) +CO2(g)
2.00 dm3 of a 3.00 M hydrochloric acid solution is added to the kettle (this is an excess).
During the reaction 1.20 dm3 of CO2 is produced (assume 250C and 1 atmosphere)

How many moles of CO2 are produced?

Answer 1

If you only want the mols, that's PV = nRT and solve for n.
P = 1 atm
V = 1.2 L
R = 0.0206
T = 273 + 250 = ?K

Answer 2

ok I checked that the answer is 0.05 but i have no idea how it's done...
Anyone can show the steps?

Answer 3

You can't plug and chug?
PV = nRT
The problem says you have 1.2 dm^23 (which is 1L) CO2 at the conditions listed.
P = 1 atm
V = 1.2 L
n = ?
R = 0.08206 L*atm/mol*K
T = 273 + 250 = 523 K
(1)(1.2) = n*0.08206*523
n = (1*1.2)/(0.08206*523)
n = 0.02796 mols which I would round to 0.028 to two significant figures. Your 0.05 number is not right (or you copied the problem wrong or looked up the wrong answer).