Question

What volume of CO2(g), measured at STP is produced if 15.2 grams of CaCO(s) is heated?

Answer 1

Volume = 3.4 L

Step-by-step explanation:

In order to calculate the volume of CO₂ produced when 15.2 g of CaCO₃ is heated, we need to first write out the balanced equation of the thermal decomposition of CaCO₃:

CaCO₃ (s) + [Heat] ⇒ CaO (s) + CO₂ (g)

Now, let's calculate the number of moles in 15.2 g CaCO₃:

mole no. =
`\mathrm{(mass)/(molar \ mass)}`

=
`(15.2)/(40.1 + 12 + (16 * 3))`

= 0.1518 moles

From the balanced equation above, we can see that the stoichiometric molar ratios of CaCO₃ and CO₂ are equal. Therefore, the number of moles of CO₂ produced is also 0.1518 moles.

Hence, from the formula for the number of moles of a gas, we can calculate the volume of CO₂:

mole no. =
`\mathrm{(Volume \ in \ L)/(22.4)}`

⇒
`0.1518 = \mathrm{(Volume)/(22.4)}`

⇒ Volume = 0.1518 × 22.4

= 3.4 L

Therefore, if 15.2 g of CaCO₃ is heated, 3.4 L of CO₂ is produced at STP.