If 4.74 g of calcium carbonate is heated, the volume of [tex]CO_2[/tex] that would be produced at STP will be 1.08 L
Volume of gas at STP
From the equation of the reaction:
[tex]CaCO_3 --- > CaO + CO_2[/tex]
The mole ratio of [tex]CaCO_3[/tex] and [tex]CO_2[/tex] is 1:1
Mole of 4.74 g [tex]CaCO_3[/tex] = 4.74/100 = 0.0474 moles
Equivalent mole of [tex]CO_2[/tex] = 0.0474 moles
1 mole of gas at STP = 22.71 L
0.0474 moles at STP = 22.71 x 0.0474 = 1.08 L
More on the volume of gas at STP can be found here: https://brainly.com/question/11676583
