Question

How many liters of CO2 are formed when 14.0g of CaCO3 react at 1.00atm and 1000K

Answer 1

11.4L

Step-by-step explanation:

The given reaction is,

`CaCO_(3)->CaO+CO_(2)`

let
`n` be the number of moles of
`CaCO_(3)`

For one mole of
`CaCO_(3)`,one mole of
`CO_(2)` is formed.

`\text{number of moles}=\frac{\text{given weight}}{\text{molar weight}}`

`\text{molar weight}` of
`CaCO_(3)`=100g

Given that
`given weight=14g`

`n=(14)/(100)`

So,number of moles of
`CO_(2)=n=0.14`

using
`PV=nRT` with
`P=\text{1 atm,}T=\text{1000K,}R={\text{0.0821 L atm }mol^(-1)K^(-1)`

`v=(nRT)/(P)=0.14* 0.0821* 1000* (1)/(1) =11.4L`