Question

In the lab, students decomposed a sample of calcium carbonate by heating it over a Bunsen burner and collected carbon dioxide according to the following equation: CaCO3(s) CaO(s) CO2(g) (a) How many mL of carbon dioxide gas were generated by the decomposition of 3.18 g of calcium carbonate at STP

Answer 1

712.32ml

Step-by-step explanation:

From the equation, we can see that one mole of limestone yielded one mole carbon iv oxide.

Now, we need to know the actual number of moles reacted. The actual number of moles reacted = mass reacted ÷ molar mass of limestone.

The molar made of limestone is = 100g/mol.

Hence, the no of moles = 3.18÷100 = 0.0318 mole

Now, we need to know the number of moles of carbon iv oxide yielded. Using the 1 to 1 mole ratio, this shows that 1 0.0318 of Carbon iv oxide was yielded.

We know 1 mole of a gas at s.t.p occupies a volume of 22.4dm^3 , hence, 0.0318 will occupy; 0.0318b× 22.4 = 0.71232dm^3

1000ml = 1dm^3

Converting the volume to ml = 0.71232 × 1000 = 712.32ml