S + O2 → SO2
Moles of S:
11/32.06 = 0.3431 mol
Moles of O2:
44/15.999 = 2.75 mol
Sulfur will be the limiting reagent.
Oxygen will be the excess reagent
2.75 - 0.3431 = 2.4071 mols of oxygen leftover
2.4071 * 15.999 = 38.511
38.511 grams of oxygen will be leftover
Answer:
This a simple stoichiometry problem using the ideal gas law.
First take the grams of ammonium carbonate and convert it to moles using its molar mass and dividing. 11.9 g/96.0932 g/mol= .12384 mol
Now use a molar conversion using the balanced equation,
1 mol (NH4)2CO3 ---> 4 mol gas formed (2 mol NH3 + 1 mol CO2 + 1 mol H2O) = .12384 x 4 = .49535 mol gas
PV=nRT
V=nRT/P= .49535mol (.08206 Lxatm/molxK) (296K)/ (1.03 atm)=11.682 L
The best explanation would be that Gases were released during the Chemical reaction, causing a loss of Mass.
