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

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

$\text{Moles of }CaCO_3=\frac{14.0g}{100g/mole}=0.14mole$

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:

$(1.00atm)\times V=(0.14mole)\times (0.0821L.atm/mole.K)\times (1000K)$

$V=11.5L$

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