Question

CaCO3(s) + 2HCl(aq) ---> CaCl2(s) + CO2(g) + H2O(ℓ)What would be the volume of CO2 (at STP) produced from the complete reaction of 10.0 grams of CaCO3?

Answer 1

The equation is balanced since the number of atoms of each element is the same on each side of the reaction.

Now, we will calculate the number of moles present in 10 grams of CaCO3. We will use the molar mass of CaCO3 for this

`\begin{gathered} lesofCaCO3=gCaCO_3*\frac{1molCaCO_3}{MolarMass,\text{gCaCO}_3} \\ MolesofCaCO3=10.0gCaCO_3*(1molCaCO_3)/(100.0869gCaCO_3)=0.1molCaCO_3 \end{gathered}`

Assuming that the rest of the reactants are in excess, 0.1 mol of CaCO3 will react and form 0.1 mol of CO2, since the ratio is 1 to 1.

Now to calculate the volume, we can apply the ideal gas law which tells us:

`PV=nR_{}T`

Where,

P is the pressure (STP) =1 atm

V is the volume in Liters

n is the number of moles = 0.1 mol CO2

R is a constant = 0.08206 (atm L)/(mol K)

T is the temperature (STP)= 273.15K

Now, we clear the volume and write the known values:

`\begin{gathered} V=\frac{nR_{}T}{P} \\ V=\frac{0.1\text{mol}*0.08206(atm.L)/(mol.K)*273.15K}{1atm} \\ V=2.2L \end{gathered}`

The volume of CO2(at STP) produced would be 2.2L