Question

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

Answer 1

Answer: 11.5

Step-by-step explanation:

Answer 2

Answer : The volume of
`CO_2` formed are 11.5 liters.

Explanation :

First we have to calculate the moles of
`CaCO_3`

`\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}`

Molar mass of
`CaCO_3` = 100 g/mole

`\text{Moles of }CaCO_3=(14.0g)/(100g/mole)=0.14mole`

The balanced chemical reaction will be:

`CaCO_3\rightarrow CaO+CO_2`

From the balanced chemical reaction we conclude that,

As, 1 mole of
`CaCO_3` react to give 1 mole pf
`CO_2`

So, 0.14 mole of
`CaCO_3` react to give 0.14 mole pf
`CO_2`

Thus, the moles of
`CO_2` gas = 0.14 mole

Now we have to calculate the volume of
`CO_2`.

Using ideal gas equation:

`PV=nRT`

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 1000 K

R = gas constant = 0.0821 L.atm/mole.K

n = moles of gas = 0.14 mole

Now put all the given values in the ideal gas equation, we get:

`(1.00atm)* V=(0.14mole)* (0.0821L.atm/mole.K)* (1000K)`

`V=11.5L`

Therefore, the volume of
`CO_2` formed are 11.5 liters.