How many liters of carbon dioxide can be obtained from the reaction of 906 grams of calcium carbonate with hydrochloric acid?

Answer 1

Assume Standard temp. and pressure. Balanced equation.

CaCO3 + 2HCl -> CaCl2 + CO2 + H2O

906 grams CaCO3 ( 1 mole CaCO3/100.09 grams)(1 mole CO2/1 mole CaCO3)

= 9.05185 moles CO2 produced

Now we assume reaction is at STP ( as they usually are ) and use ideal gas law.

PV = nRT

(1 atm)(V) = (9.05185 moles)(0.0806 L*atm/mol*K)(298.15 K)

volume of CO2 in liters is;

217.52 Liters of CO2 produced in this reaction.

seems reasonable with those masses

Answer 2

To determine the number of liters of carbon dioxide produced in this reaction, we need the balanced equation and the molar mass of carbon dioxide. The balanced equation is:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

The molar mass of CO2 is 44.01 g/mol.

First, we calculate the number of moles of CaCO3:

906 g / molar mass of CaCO3 = moles of CaCO3

Using the balanced equation, we see that the stoichiometric coefficient of CO2 is 1. This means that the number of moles of CO2 produced is equal to the number of moles of CaCO3.

Finally, we convert moles of CO2 to liters using the ideal gas law:

moles of CO2 x 22.4 L/mol = liters of CO2.

Therefore, the number of liters of CO2 produced from 906 grams of CaCO3 can be calculated as follows:

liters of CO2 = (906 g / molar mass of CaCO3) x 22.4 L/mol