When heated, calcium carbonate decomposes to yield calcium oxide and carbon dioxide gas via the reaction



How many grams of calcium carbonate are needed to produce 49.0 of carbon dioxide at STP?

balance the equation

CaCO3>> CaO + CO2

so for every mole of CO2, you need one mole of calcium carbonate.

How many moles of CO2 is 49grams?

111.3636gm of CaCo3

To find out how many grams of calcium carbonate are needed to produce 49.0 grams of carbon dioxide, we need to use stoichiometry. Stoichiometry is a way to relate the quantities of substances involved in a chemical reaction.

First, we need to know the balanced chemical equation for the reaction:

CaCO₃ → CaO + CO₂

According to the balanced equation, 1 mole of calcium carbonate (CaCO₃) produces 1 mole of carbon dioxide (CO₂).

To convert between grams and moles, we need to know the molar mass of calcium carbonate and carbon dioxide.

The molar mass of CaCO₃ is calculated by adding up the atomic masses of calcium (Ca), carbon (C), and three oxygen (O) atoms.

CaCO₃: (1 × Ca) + (1 × C) + (3 × O) = 40.08 g/mol + 12.01 g/mol + (3 × 16.00 g/mol) = 100.09 g/mol

The molar mass of CO₂ is calculated by adding up the atomic masses of carbon (C) and two oxygen (O) atoms.

CO₂: (1 × C) + (2 × O) = 12.01 g/mol + (2 × 16.00 g/mol) = 44.01 g/mol

Now, we can use the molar masses and stoichiometry to convert grams of carbon dioxide to grams of calcium carbonate.

First, find the number of moles of carbon dioxide:
49.0 g CO₂ × (1 mol CO₂ / 44.01 g CO₂) = 1.11 mol CO₂

Since the ratio between moles of calcium carbonate and moles of carbon dioxide is 1:1, we have 1.11 moles of calcium carbonate.

Finally, convert moles of calcium carbonate to grams:
1.11 mol CaCO₃ × (100.09 g CaCO₃ / 1 mol CaCO₃) = 111 g CaCO₃

Therefore, 111 grams of calcium carbonate are needed to produce 49.0 grams of carbon dioxide at STP.