What volume, in mL, of carbon dioxide gas is produced at STP by the decomposition of 0.242 g calcium carbonate (the products are solid calcium oxide and carbon dioxide gas). Report your answer to 4 sig figs

54.21 mL.

Explanation:

We'll begin by calculating the number of mole in 0.242 g calcium carbonate, CaCO3.

This is illustrated below:

Mass of CaCO3 = 0.242 g

Molar mass of CaCO3 = 40 + 12 +(16x3) = 40+ 12 + 48 = 100 g/mol

Mole of CaCO3 =?

Mole = mass /Molar mass

Mole of CaCO3 = 0.242/100

Mole of CaCO3 = 2.42×10¯³ mole.

Next, we shall write the balanced equation for the reaction. This is given below:

CaCO3 —> CaO + CO2

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole CaO and 1 mole of CO2.

Next, we shall determine the number of mole of CO2 produced from the reaction.

This can be obtained as follow:

From the balanced equation above,

1 mole of CaCO3 decomposed to produce 1 mole of CO2.

Therefore,

2.42×10¯³ mole of CaCO3 will also decompose to produce 2.42×10¯³ mole of CO2.

Therefore, 2.42×10¯³ mole of CO2 were obtained from the reaction.

Finally, we shall determine volume occupied by 2.42×10¯³ mole of CO2.

This can be obtained as follow:

1 mole of CO2 occupies 22400 mL at STP.

Therefore, 2.42×10¯³ mole of CO2 will occupy = 2.42×10¯³ x 22400 = 54.21 mL

Therefore, 54.21 mL of CO2 were obtained from the reaction.