How many grams of calcium carbonate are decomposed to produce 11.2 l of carbon dioxide at STP?
First, we need to use the ideal gas law to convert 11.2 L of carbon dioxide at STP to moles:
PV = nRT
(1 atm) (11.2 L) = n (0.0821 L atm/mol K) (273 K)
n = 0.486 mol
Now, we use the balanced chemical equation for the decomposition of calcium carbonate:
CaCO3(s) → CaO(s) + CO2(g)
1 mole of CaCO3 produces 1 mole of CO2. Therefore, the number of moles of CaCO3 decomposed is also 0.486 mol.
Finally, we use the molar mass of CaCO3 to convert mol to g:
0.486 mol CaCO3 x 100.1 g/mol = 48.7 g
Therefore, 48.7 g of calcium carbonate are decomposed to produce 11.2 L of carbon dioxide at STP.