Question

According to the following reaction: ___ CaCO3(s)→___ CO2(g) + ___ CaO(s)a. How many grams of calcium carbonate will be needed to form 4.29 liters of carbon dioxide at STP?

Answer 1

Step 1

The reaction must be written, completed, and balanced:

1 CaCO3(s) → 1 CO2(g) +1 CaO(s)

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Step 2

It is known that at STP conditions:

1 mole of gas (CO2 in this case) = 22.4 L of CO2

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Step 3

Information needed:

The molar mass of calcium carbonate (CaCO3) = 100 g/mol

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Step 4

By stoichiometry:

1 mole of CO2 at STP = 22.4 L

1 mole of CaCO3 = 100 g

1 CaCO3(s) → 1 CO2(g) +1 CaO(s)

Therefore,

100 g CaCO3 ----------- 22.4 L CO2

X ----------- 4.29 L CO2

X = 4.29 L CO2 x 100 g CaCO3/22.4 L CO2 = 19.2 g CaCO3

Answer: 19.2 g CaCO3