Answer:
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.
