Question

Calculate the pressure at STP with 10.0 L vessel at the end of the reaction

1.0L of HCL (35% with a density of 1.28 g/ ml ) and 1.0kg of calcium bicarbonate ( this has a 4.0% impurity) the reaction yields 97%

Answer 1

First of all, we need to know the balanced reaction that is involved in the process. It is:

2 HCl + Ca(HCO₃)₂ → CaCl₂ + 2 CO₂ + 2 H₂O

Where all the species involved are either solids or liquids, excep for CO₂, which is a gas and will be responsible for the increase of pressure during the reaction. We now have to work a little with the stoichiometry of the reaction. Taking into account the amounts of reactants, we can calculate the number of moles they represent:

For HCl:

1.0 L reacts, and the concentration is 35% (in weight). We thus have to know the mass of HCl solution added, and for this, we use the density. By definition:

Thus:

There are 1280 grams of HCl solution involved in the reaction, and 35% of this mass is pure HCl, thus, the mass of pure HCl is given by:

Which can be converted to number of moles dividing by the molar mass of HCl:

For Ca(HCO₃)₂:

There is 1 kg of reactant, but with a 4% impurty. Thus, only 96% of the total mass is pure calcium bicarbonate; that is: 960 g.

These can be converted to moles using:

We now know the available amount of moles of reactants. The next step is to determine which of them is the limiting reactant, which will be completely consumed. A simple way of determining the limiting reactant is to divide the number of moles that react by the coefficient of the reactant in the balanced chemical equation. Thus:

The reactant that yields the lowest result after this operation is the limiting reactant. In this case, calcium bicarbonate is the limiting reactant.

The next step is to determine the amount of CO₂ that will be generated when 5.92 moles of calcium bicarbonate react. The stoichiometry of the reaction indicates that per each mole of calcium bicarbonate, two moles of CO₂ are generated. Thus, in our case, if the reaction had a 100% yield, we would get 2*5.92 = 11.84 moles of CO₂. But since the yield is not 100%, but 97%, this value is a little lower:

Thus, a 97% yields results in 11.48 moles of carbon dioxide.

The last step is to determine the pressure generated by the generated CO₂ in the 10 L vessel. Since the volume occupied by the other substances involved in the reaction (solids and liquids), we will consider only the volume of the gas as relevant. To determine this pressure, we can assume the gas is ideal and use the ideal gases equation:

p*V = n*R*T

Which can be re-arranged to:

P=n*R*T/V

Applying our values:

The final pressure in the vessel is then 28.1 Atm